[SOLVED] 代写 data structure ocaml 1 Overview

1 Overview
CS 320: Language Interpreter Design
Part 1 Due: October 29, at 11:59pm Part 2 Due: November 12, at 11:59pm Part 3 Due: November 26, at 11:59pm
The goal of this project is to understand and build an interpreter for a small, OCaml-like, stack- based bytecode language. You will be implementing this interpreter in OCaml, like the previous assignments. The project is broken down into three parts. Part 1 is defined in Section 4, Part 2 is defined in Section 5, and Part 3 is defined in Section 6. Each part is worth 80 points.
You will submit a file named interpreter.ml which contains a function, interpreter, with the following type signature:
val interpreter : string -> string -> unit
If your program does not match the type signature, it will not compile on Gradescope and you will receive 0 points. You may, however, have helper functions defined outside of interpreter—the grader is only explicitly concerned with the type of interpreter.
You must submit a solution for each part and each part is graded individually. Late submissions will not be accepted and will be given a score of 0. Test cases sample will also be provided on Piazza for you to test your code locally. These will not be exhaustive, so you are highly encouraged to write your own tests to check your interpreter against all the functionality described in this document.
2 Functionality
Given the following function header:
let interpreter (input :string) (output :string ) :unit = …
input file name and output file name will be passed in as strings that represent paths to files just like in the Pangram assignment. Your function should write to the contents of the final stack your interpreter produces to the file specified by output. In the examples below, the input file is read from top to bottom and then each command is executed by your interpreter in the order it was read. It is incredibly useful to read in all of the commands into a list prior to executing them, separating input from the actual interpretation of the commands. The input file can be arbitrarily long. You may find the library function String.split_on_char to be useful for separating a string into a string list.
1

input
stack
input
stack
PushI 10 PushI 2 PushI 8 Mul
Add PushI 3 Sub Quit
input
stack
PushB PushI 7
PushI 8
PushB Pop
Sub Quit
input
stack
PushI 5 Neg PushI 10 PushI 20 Add Quit
PushI 6 PushI 2 Div Mul Quit
input
PushI 1 Quit
stack
3
1
30 -5
-1 23
3 Grammar
The following is a context free grammar for the bytecode language you will be implementing. Terminal symbols are identified by monospace font, and nonterminal symbols are identified by italic font. Anything enclosed in [brackets] denotes an optional character (zero or one occurrences). The form ‘( set1 | set2 | setn)’ means a choice of one character from any one of the n sets. A set enclosed in {braces means zero or more occurrences}.
The set digit is the set of digits {0,1,2,3,4,5,6,7,8,9}, letter is the set of all characters in the English alphabet (lowercase and uppercase), and ASCII is the ASCII character set. The set sim- pleASCII is ASCII without quotation marks and the backslash character. Do note that this neces- sarily implies that escape sequences will not need to be handled in your code.
3.1 Constants
const ::= int | bool | error | string | name | unit int ::= [−] digit { digit }
bool ::= |
error ::=
unit ::=
string ::= “simpleASCII { simpleASCII }” simpleASCII ::= ASCII {‘’, ‘”‘} name ::= {_} letter {letter | digit | _}
3.2 Programs
prog ::= com {com} Quit
com ::= PushI const | PushB const | PushS const | PushN const | Push const | Add | Sub | Mul | Div | Rem | Neg | Swap | Pop | Concat | And | Or | Not | LessThan | Equal | If | Bind |
Begin com {com} End | funBind com {com} [ Return ] FunEnd | Call
funBind ::= (Fun | InOutFun) name1 name2
2

4 Part 1: Basic Computation
Due Date: October 27, at 11:59pm
Your interpreter should be able to handle the following commands:
4.1 Push
4.1.1 pushing Integers to the Stack
PushI num
where num is an integer, possibly with a ’-’ suggesting a negative value. Here ’-0’ should be regarded
as ’0’. Entering this expression will simply Push num onto the stack. For example,
If num is not an integer, only Push the error literal ( ) onto the stack instead of pushing num. For example,
4.1.2 pushing Strings to the Stack
PushS string
where string is a string literal consisting of a sequence of characters enclosed in double quotation
marks, as in “this is a string”. Executing this command would Push the string onto the stack:
Spaces are preserved in the string, i.e. any preceding or trailing whitespace must be kept inside the string that is Pushed to the stack:
You can assume that the string value would always be legal and not contain quotations or escape sequences within the string itself, i.e. neither double quotes nor backslashes will appear inside a string.
If string is not string, only Push the error literal ( ) onto the stack instead of pushing string.
input
stack
PushI 5 PushI -0
0 5
input
stack
PushI 5 PushI 2.5 PushI x
5
input
stack
PushS “deadpool” PushS “batman” PushS “this is a string”
this a string batman deadpool
input
stack
PushS ” deadp ool ” PushS “this is a string ”
this␣is␣a␣string␣␣ ␣deadp␣ool␣
3

4.2 pushing Names to the Stack
PushN name
where name consists of a sequence of letters, digits, and underscores, starting with a letter or
underscore.
1. example
2. example
→→
→→
input
PushN a PushI 13
stack
13 a
stack
a
stack
3 __name1__
input
PushN __name1__ PushI 3
stack
__name1__
If name does not conform to previously mentioned specifications, only Push the error literal ( ) onto the stack instead of pushing name.
To bind ‘a’ to the value 13 and __name1__ to the value 3, we will use the ‘Bind’ operation which we will see later (Section 5.7) You can assume that name will not contain any illegal tokens— no commas, quotation marks, etc. It will always be a sequence of letters, digits, and underscores, starting with a letter (uppercase or lowercase) or an underscore.
4.3 boolean
PushB bool
There are two kinds of boolean literals: and . Your interpreter should Push the
corresponding value onto the stack. For example,
If bool is not a boolean, only Push the error literal ( ) onto the stack instead of pushing bool.
4.4 error and unit
Push Push
Similar with boolean literals, pushing an error literal or unit literal will Push or onto the stack, respectively. If anything except these two literals are passed to Push (without the suffix), Push onto the stack.
input
stack
PushI 5
PushB 5
4

input
Push Push Push Push Push Quit
stack

4.5 Pop
The command Pop removes the top value from the stack. If the stack is empty, an error literal ( ) will be Pushed onto the stack. For example,
→→→
4.6 Add
The command Add refers to integer addition. Since this is a binary operator, it consumes the top two values in the stack, calculates the sum and Pushes the result back to the stack. If one of the following cases occurs, which means there is an error, any values popped out from the stack should be Pushed back in the same order, then a value should also be Pushed onto the stack:
• not all top two values are integer numbers • only one value in the stack
• stack is empty
for example, the following non-error case:
→→
Alternately, if there is only one number in the stack and we use Add, an error will occur. Then 5 should be Pushed back as well as
→→
→
input
PushI 5 Pop Pop
stack
5
stack
input

input
PushI 5 PushI 8 Add
stack
8 5
stack
13
input
PushI 5 Add
stack
5
stack
5
5

4.7 Sub
The command Sub refers to integer subtraction. It is a binary operator and works in the following way:
• if top two elements in the stack are integer numbers, pop the top element(y) and the next element(x), subtract x from y, and Push the result y-x back onto the stack
• if the top two elements in the stack are not all integer numbers, Push them back in the same order and Push onto the stack
• if there is only one element in the stack, Push it back and Push onto the stack
• if the stack is empty, Push onto the stack
for example, the following non-error case:
→→
Alternately, if one of the top two values in the stack is not a numeric number when Sub is used, an error will occur. Then 5 and should be Pushed back as well as
→→→
4.8 Mul
The command Mul refers to integer multiplication. It is a binary operator and works in the following way:
• if top two elements in the stack are integer numbers, pop the top element(y) and the next element(x), multiply x by y, and Push the result x*y back onto the stack
• if the top two elements in the stack are not all integer numbers, Push them back in the same order and Push onto the stack
• if there is only one element in the stack, Push it back and Push onto the stack
• if the stack is empty, Push onto the stack For example, the following non-error case:
→→
input
PushI 5 PushI 8 Sub
stack
8 5
stack
3
input
PushI 5
PushB Sub
stack
5
stack
5
stack
5
input
PushI 5 PushI 8 Mul
stack
8 5
stack
40
6

Alternately, if the stack empty when Mul is executed, an error will occur and should be Pushed onto the stack:
→→
4.9 Div
The command Div refers to integer division. It is a binary operator and works in the following way:
• if top two elements in the stack are integer numbers, pop the top element(y) and the next
element(x), divide y by x, and push the result y back onto the stack x
• if top two elements in the stack are integer numbers but x equals to 0, Push them back in the same order and Push onto the stack
• if the top two elements in the stack are not all integer numbers, Push them back in the same order and Push onto the stack
• if there is only one element in the stack, Push it back and Push onto the stack
• if the stack is empty, Push onto the stack
For example, the following non-error case:
→→
Alternately, if the second top element in the stack equals to 0, there will be an error if Div is executed. In such situations 0 and 5 should be Pushed back onto the stack as well as
→→
4.10 Rem
The command Rem refers to the remainder of integer division. It is a binary operator and works in the following way:
• if top two elements in the stack are integer numbers, pop the top element(y) and the next element(x), calculate the remainder of y , and Push the result back onto the stack
x
• if top two elements in the stack are integer numbers but x equals to 0, Push them back in the same order and Push onto the stack
• if the top two elements in the stack are not all integer numbers, Push them back and Push onto the stack
input
Mul
stack
stack

input
PushI 5 PushI 8 Div
stack
8 5
stack
1
input
PushI 0 PushI 5 Div
stack
5
0
stack
5 0
7

• if there is only one element in the stack, Push it back and Push onto the stack • if the stack is empty, Push onto the stack
For example, the following non-error case:
→→
Alternately, if one of the top two elements in the stack is not an integer, an error will occur if Rem is executed. If this occurs the top two elements should be Pushed back onto the stack as well as . For example:
→→
4.11 Neg
The command Neg is to calculate the negation of an integer (negation of 0 should still be 0). It is unary therefore consumes only the top element from the stack, calculate its negation and Push the result back. A value will be Pushed onto the stack if:
• the top element is not an integer, Push the top element back and Push • the stack is empty, Push onto the stack
For example, the following non-error case:
→→
Alternately, if the value on top of the stack is not an integer, when Neg is used, that value should be Pushed back onto the stack as well as . For example:
→→→
input
PushI 5 PushI 8 Rem
stack
8 5
stack
3
input
PushI 5
PushB Rem
stack
5
stack
5
input
PushI 5 Neg
stack
5
stack
-5
input
PushI 5
Neg
PushB Neg
stack
-5
stack
-5
stack
-5
8

4.12 Swap
The command Swap interchanges the top two elements in the stack, meaning that the first element becomes the second and the second becomes the first. A value will be Pushed onto the stack if:
• there is only one element in the stack, Push the element back and Push • the stack is empty, Push onto the stack
For example, the following non-error case:
→→→
Alternately, if there is only one element in the stack when Swap is used, an error will occur and should be Pushed onto the stack. Now we have two elements in the stack (5 and ), therefore the second Swap will interchange the two elements:
→→→
4.13 Quit
The command Quit causes the interpreter to stop. Then the whole stack should be printed out to the output file that is specified as the second argument to the interpreter function.
input
PushI 5
PushI 8
PushB Swap
stack
8
5
stack
8 5
stack
8 5
input
PushI 5 Swap Swap
stack
5
stack
5
stack
5
9

5 Part 2: Variables and Scope
Due date: November 10, at 11:59pm
In part 2 of the interpreter you will be expanding the types of computation you will be able to perform, adding support for immutable variables and structures for expressing scope.
5.1 Concat
The Concat command computes the concatenation of the top two elements in the stack and Pushes the result onto the stack. The top two values of the stack — x and y — are popped off and the result is the string x concatenated onto y.
will be Pushed onto the stack if:
• there is only one element in the stack, Push the element back and Push
• the stack is empty, Push onto the stack
• if either of the top two elements are not strings, Push the elements back onto the stack, and then Push
– Hint: Recall that names and strings are different. For example:
input
PushS “world!” PushS “hello ” Concat
stack
hello world!
stack
world!
stack
hello world!
Consider another example:
→→→
→→→
input
PushN Scott PushS “Michael” Concat
stack
Michael Scott
stack
Michael Scott
stack
Scott
Note that strings can contain spaces, punctuation marks, and other special characters. You may assume that strings only contain ASCII characters and have no escape sequences, e.g.
and t.
5.2 And
The command And performs the logical conjunction of the top two elements in the stack and Pushes the result (a single value) onto the stack.
will be Pushed onto the stack if:
• there is only one element in the stack, Push the element back and Push
• the stack is empty, Push onto the stack
• if either of the top two elements are not booleans, Push back the elements and Push
10

For example:
Consider another example:
5.3 Or
→→→
→→
input
PushB PushB And
stack
stack
stack

input
PushB And
stack
stack
The command Or performs the logical disjunction of the top two elements in the stack and Pushes the result (a single value) onto the stack.
will be Pushed onto the stack if:
• there is only one element in the stack, Push the element back and Push
• the stack is empty, Push onto the stack
• if either of the top two elements are not booleans, Push back the elements and Push
For example:
Consider another example:
5.4 Not
→→→
→→→
input
PushB PushB Or
stack
stack
stack
input
PushB PushS “khaleesi” Or
stack
khaleesi
stack
khaleesi
stack

The command Not performs the logical negation of the top element in the stack and Pushes the result (a single value) onto the stack. Since the operator is unary, it only consumes the top value from the stack. The value will be Pushed onto the stack if:
• the stack is empty, Push onto the stack
• if the top element is not a boolean, Push back the element and Push
11

For example:
Consider another example:
5.5 Equal
→→
→→
input
PushB Not
stack
stack

input
PushI 3 Not
stack
3
stack
3
The command Equal refers to numeric equality (so you are not supporting string comparisons). This operator consumes the top two values on the stack and Pushes the result (a single boolean value) onto the stack. The value will be Pushed onto the stack if:
• there is only one element in the stack, Push the element back and Push
• the stack is empty, Push onto the stack
• if either of the top two elements are not integers, Push back the elements and Push
For example:
Consider another example:
5.6 LessThan
→→→
→→→
input
PushI 7 PushI 7 Equal
stack
7 7
stack
7
stack
input
PushI 8 PushI 9.5 Equal
stack
8
stack
8
stack
8
The command LessThan refers to numeric less than ordering. This operator consumes the top two values on the stack and Pushes the result (a single boolean value) onto the stack. The value will be Pushed onto the stack if:
• there is only one element in the stack, Push the element back and Push
• the stack is empty, Push onto the stack
• if either of the top two elements aren’t integers, Push back the elements and Push 12

For example:
5.7 Bind
→→→
input
PushI 7 PushI 8 LessThan
stack
8 7
stack
7
stack

The Bind command binds a name to a value. It is evaluated by popping two values from the stack. The first value popped must be a name (see section 4.2 for details on what constitutes a ’name’). The name is bound to the value (the second thing popped off the stack). The value can be any of the following:
• An integer • A string
• A boolean •
• The value of a name that has been previously bound
The name value binding is stored in an environment data structure. The result of a Bind operation is which is Pushed onto the stack. The value will be Pushed onto the stack if:
• •
5.7.1
5.7.2
we are trying to bind an identifier to an unbound identifier, in which case all elements popped must be Pushed back before pushing onto the stack.
the stack is empty, Push onto the stack. Example 1
→→→
Example 2
→→→→→→
input
PushI 3 PushN a Bind
stack
a 3
stack
3
stack

input
PushI 7 PushN sum1 Bind
PushI 5 PushN sum2 Bind
stack
sum2
5
stack
sum1 7
stack
5
stack

stack
7
stack

13

You can use bindings to hold values which could be later retrieved and used by functionalities you already implemented. For instance, in the example below, an addition on a and name1 would add 13 + 3 and Push the result 16 onto the stack.
This, in effect, allows names to be in place of proper constants in all the operations we’ve seen so far. Take for example, when you encounter a name in an Add operation, you should retrieve the value the name is bound to, if any. Then if the value the name is bound to has the proper type, you can perform the operation.
5.7.3 Example 3
→→→→→→→
→→
Notice how we can substitute a constant for a bound name and the commands work as we expect. The idea is that when we encounter names in a command, we resolve the name to the value it’s bound to, and then use that value in the operation.
input
PushI 13 PushN a Bind
PushI 3 PushN name1 Bind
PushN a PushN name1 Add
stack
name1 3
stack
a 13
stack
3
stack

stack
13
stack

stack
name1 a
stack
a
stack
16
14

5.8 Example 4
input
PushI 5 PushN a Bind
Pop
PushI 3 PushN a Add
PushS “str” PushN b Bind
Pop PushI 10 PushN b Sub Quit
stack
b
10
8
You can see that the Add operation completes, because a is bound to an integer (5, specifically). The Sub operation fails because b is bound to a string, and thus does not type check. While performing operations, if a name has no binding or it evaluates to an improper type, Push onto the stack, in which case all elements popped must be Pushed back before pushing onto the stack.
5.9 Example 5
Bindings can be overwritten, for instance:
→
input
PushI 9 PushN a Bind PushI 10 PushN a Bind
Here, the second Bind updates the value of a to 10. Common Questions
(a) What values can _name_ be bound to?
_name_ can be bound to integers, booleans, strings, and also previously bound val- ues. For example,
15

input
PushB PushN a
Bind
1)
would bind a to 2)
would result in Bind producing an because a CANNOT be bound to
3)
would bind a to 7 and b to
4)
would bind b to 8 and would bind a to the VALUE OF b which is 8.
5)
would result in an because you are trying to bind b to an unbound variable a. (b) How can we bind identifiers to previously bound values?
input
PushI 7.5 PushN a Bind
input
Begin PushI 7 PushN a Bind End PushN b Bind
input
PushI 8 PushN b Bind PushN b PushN a Bind
input
PushN b PushN a Bind
16

input
PushI 7 PushN a Bind PushN a PushN b Bind
The first Bind binds the value of a to 7. The second Bind statement would result in the name b getting bound to the VALUE of a—which is 7. This is how we can bind identifiers to previously bound values. Note that we are not binding b to a—we are binding it to the VALUE of a.
(c) Can we have something like this?
Yes. In this case a is not bound to any value yet, and the stack contains:
If we had:
The stack would be:
(d) Can we Push the same _name_ twice to the stack? For instance, what would be the result of the following:
input
PushI 15 PushN a PushN a
stack
a a 15
input
PushI 15 PushN a Bind PushN a
stack
a
17

input
PushN a PushN a Quit
This would result in the following stack output:
Yes, you can push the same _name_ twice to the stack. Consider binding it this way:
This would result in
→ as we cannot bind a unbound name a to a name a a → as a result of pushing the second a to the stack
a → as a result of pushing the first a to the stack
2 → as a result of pushing the first 2 to the stack
(e) Output of the following code:
stack
a a
input
PushI 2 PushN a PushN a Bind
input
PushI 9 PushN a Bind PushI 10 PushN a Bind
This would result in the following stack output:
would result in
→ as a result of second Bind → as a result of first Bind
18

5.10 If
The If command pops three values off the stack: x, y and z. The third value popped (z, in this case) must always be a boolean. If z is , executing the If command will Push y back onto the stack, and if z is , executing the If will Push x back onto the stack.
will be pushed onto the stack if:
• the third value is not a boolean, all elements (x, y, and z) should be Pushed back onto the stack before pushing onto the stack.
• the stack is empty, push onto the stack
• there are less than 3 values on the stack, in which case all elements popped must be pushed
back before pushing onto the stack. For example:
→→→→
Common Questions
(a) What values can ‘If’ take?
The result of executing a ‘If’ can be an integer or boolean or string or or For instance,
1)
the result of If would be oracle
2)
the result of If would be
(b) What is the result of executing the following: 19
input
PushB PushI 9
PushI 8
If
stack
8
9 stack
9 stack
stack
9
input
PushB PushS “oracle” PushS “jive”
If
input
PushB PushI 8.9 Begin
PushI 8
PushN a Bind End
If

input
PushI 5 PushN a
Bind
Pop
PushB PushN a PushI 4
If
The stack would have a. Although the value of a is bound to 5, we only resolve the name to the value if we need to perform computation. (For ‘If’, the only value needed for computation is a boolean.)
5.11 Begin…End
Begin…End limits the scope of variables. “Begin” marks the beginning of a new environment— which is basically a sequence of bindings. The result of the Begin…End is the last stack frame of the Begin. Begin…End can contain any number of operations but it will always result in a stack frame that is strictly larger than the stack prior to the Begin.
Trying to access an element that is not in scope of the Begin…End block would Push on the stack. Begin…End blocks can also be nested.
For example,
20

input
Begin
PushI 13 PushN c Bind
Begin
PushI 3 PushN a Bind
PushN a PushN c Add
End
Begin
PushS “ron” PushN b Bind
End End
21

Original Stack
1st Begin Expression
→→→
2nd Begin Expression
→→→→→
→→
3rd Begin Expression
→→→
In the above example, the first Begin statement creates an empty environment (environment 1), then the name c is bound to 13. The result of this Bind is a on the stack and a name value pair in the environment. The second Begin statement creates a second empty environment. Name a is bound here. To Add a and c, these names are first looked up for their values in the current environment. If the value isn’t found in the current environment, it is searched in the outer environment. Here, c is found from environment 1. The sum is Pushed to the stack. A third environment is created with one binding ‘b’. The second last end is to end the scope of environment 3 and the last end statement is to end the scope of environment 1. You can assume that the stack is left with at least 1 item after the execution of any Begin…End block.
Common Questions
(a) What would be the output of running the following: 22
stack
13 c
stack
c
stack

stack
c
a
stack
3
a
stack
a
stack
16
stack
a
stack

stack
16
stack
ron
b
16
stack
b
16
stack
16
stack

input
PushI 1 Begin PushI 2 PushI 3 PushI 4 End PushI 5
This would result in the stack:
Explanation: After the Begin…End is executed the last frame is returned—which is why we have 4 on the stack.
(b) What would be the result of executing the following:
stack
5 4 1
input
Begin PushI 7.2 PushN a1 Bind
End
Quit
7.2 can’t be Pushed to the stack and a1 cannot be bound to so, the result would be
(c) What would be the output of running the following code:
The stack output would be:
input
Begin PushI 3 PushI 10 End Add Quit
stack
10
23

6 Part 3: Functions Due date: November 24, at 11:59pm
6.1 Function declaration and Call
Fun name1 name2
Denotes a function declaration, i.e. the start of a function called name1, which has one formal parameter name2. The expressions that follow comprise the function body. The function body is terminated with a special keyword FunEnd. Note, name1 and name2 can be any valid name, but will never be any of the keywords in our language (e.g. Add, Push, Pop, Fun, FunEnd, etc.). Also the function name and argument name cannot be the same.
denotes the end of a function body.
FunEnd
PushN funName Push arg Call
Denotes applying the function funName to the actual parameter arg. Do note that Push arg can leverage any form of the Push command, i.e. PushI, PushB, PushN, PushS, Push. When Call is evaluated, it will apply the function funName to arg and Pop both funName and arg from the stack. arg can either be a name (this includes function names), an integer, a string, a boolean, or .
When the interpreter encounters a function declaration expression it should being construction a closure. A closure will consist of (1) an environment, (2) the code for the function (the expressions between the function declaration and FunEnd), and (3) the name of the formal parameter. The value should be Pushed to the stack once the function declaration is evaluated and the closure created and bound to the function name in the environment.
1. The environment for the closure will be a copy of the current environment. (Challenge: if you would like to optimize your closure representation you do not need the entire environment, just the bindings of the variables used inside the function that are not defined inside the function and are not the formal parameter).
2. To compute the code for the function, you should copy all the expressions in order starting with the first expressions after the function declaration up to, but not including, the FunEnd.
3. In the current environment you should create a binding between the function name and its closure.
When a function is called, you should first check to see if there is a binding in the current environment, which maps funName to a closure. If one does not exist, Push onto the stack. You should then check to see if the current environment contains a binding for arg if it is a name instead of a value. If it does not, then you should Push onto the stack. If arg is an you should Push onto the stack.
If both funName and arg have appropriate bindings, or arg is a valid value, then the Call to the function can proceed. To do this, Push the environment stored in the closure onto the stack. To this environment add a binding between the formal parameter 1 and the value of the actual
1you will extract the formal parameter from the closure 24

parameter (i.e. the argument). Note that if arg is a name, then it must have a binding in the environment at the point of the Call 2. You should then save the current stack and create a new stack that will be used for the execution of the function 3. Next retrieve the code for the function and begin executing the expressions. The function completes once the last expression in code for the function is executed. When this happens, you should restore the environment to the environment that existed prior to the function Call 4. The stack should also be restored to what the stack was at the point of the Call 5. Once the environment has been restored, execution should resume with the expression that follows the Call.
6.2 Return
Functions can return values by using a Return expression. Since functions themselves are values (a closure), functions can take other functions as arguments and can return functions. When a Return expression is evaluated, the function stops execution. When this happens you should restore the environment to the environment that existed prior to the function Call, just like if the function completed by executing the last expression in the function’s code. The stack should also be restored to what the stack was at the point of the Call. Additionally, you should Push the last stack frame the function Pushed onto the restored stack (the stack at the point of the Call).
Please note that background color and indentation is used only to improve readability. Closure would consist of code within colored background.
6.3 Examples
6.3.1 Example 1
input
Fun identity x PushN x Return FunEnd
PushN identity PushI 1
Call
Quit
stack
1
1 → return value of calling identity and passing in x as an argument → result of declaring identity
2this is the current environment before you Push the closure’s environment
3hint: you may want to implement the stack as a stack of stacks to handled nested function calls and recursion, much like implementing the environment as a stack of maps
4hint: if you are implementing your environment as a stack of local environments, this will entail popping off the top environment
5hint: if you implemented your stack as a stack of stacks, this only requires popping off the top stack to restore the stack to what it was prior to the Call
→
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6.3.2 Example 2
input
Fun identity x PushN x Return FunEnd
PushN identity PushI 1.2
Call
Quit
stack
identity
→ error as a result of calling a function with error as the actual parameter → result of pushing 1.2
identity → Push of identity
→ result of declaring identity
6.3.3 Example 3
→
1 → return value of calling identity and passing in x as an argument → result of binding x
→ result of declaring identity
→
input
Fun identity x PushN x Return FunEnd
PushI 1
PushN x
Bind
PushN identity PushN x
Call Quit
stack
1
26

6.3.4 Example 4
input
PushI 3 PushN x Bind
Fun addX arg PushN x PushN arg Add
Return FunEnd
PushI 5 PushN x Bind
PushI 3 PushN a Bind
PushN addX PushN a Call
Quit
stack
6
6 → result of function call
→ result of third binding → result of second binding → result of function declaration → result of first binding
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→

6.3.5 Example 5
input
Fun stop arg PushI 1 Return FunEnd
Fun factorial arg PushN arg PushI 1
Sub
PushI 1
PushN arg Equal
PushN factorial PushN stop
If
Swap
Call PushN arg Mul Return FunEnd
PushN factorial PushI 3
Call
Quit
stack
6
6 → value returned from factorial → declaration of factorial → declaration of stop
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→

6.3.6 Example 6
input
Fun add1 x PushN x PushI 1 Add Return FunEnd
PushI 2 PushN z Bind
Fun twiceZ y PushN y PushN z Call
PushN y PushN z Call PushN y PushN z Call Add Return FunEnd
PushN twiceZ PushN add1 Call
Quit
stack
6
→
6 → return of calling twiceZ and passing add1 as an argument → declaration of twiceZ
→ binding of z
→ declaration of the add1 function
6.4 Functions and Begin
Functions can be declared inside a Begin expression. Much like the lifetime of a variable binding, the binding of a function obeys the same rules. Since Begin introduces a stack of environments, the closure should also take this into account. The easiest way to implement this is for the closure to store the stack of environments present at the declaration of the function. (Note: you can create a more optimal implementation by only storing the bindings of the free variables used in the function—to do this you would look up each free variable in the current environment and add a binding from the free variable to the value in the environment stored in the closure)
(please note background color is used only to improve readability):
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6.4.1 Example 1
input
Begin
Fun identity x PushN x Return FunEnd
end
PushN identity PushI 1
Call
Quit
stack
1 identity
→ error since identity is not bound in the environment
1 → Push of 1
identity → Push of identity
→ result of declaring identity, this is the result of the Begin expression
6.4.2 Example 2
→
1 → return value of calling identity and passing in x as an argument → result of declaring identity
→
input
Fun identity x
Begin PushN x end
Return FunEnd
PushN identity PushI 1
Call
Quit
stack
1
30

6.4.3 Example 3
input
Fun double x
Begin PushN x PushN x Add
end
Return FunEnd
PushN double PushI 2
Call
Quit
stack
4
4 → return value of calling identity and passing in x as an argument → result of declaring identity
6.4.4 Example 4
→
input
PushI 5 PushN y Bind
Begin PushI 7 PushN y Bind
Fun addY x
Begin PushN x PushN y Add
end
Return FunEnd
PushN addY PushI 2
Call
end
Quit
stack
9
31
→

9 → return value of calling identity and passing in 2 as an argument → result of binding y to 5
6.5 In/Out Functions
Our language will also support in/out parameters for specially denoted functions. Instead of using the Fun keyword, functions that have in/out parameters are declared using the InOutFun keyword. In/out functions behave just like regular functions and all the rules defined for functions apply. In addition, when an in/out function returns, the value bound to the formal parameter is bound to the actual parameter in the environment after the Call.
In/out functions should have a similar implementation to regular functions. To this implementa- tion you should add an additional operation when the function returns. In addition to restoring the environment at the Call site, the Return will do a look up of formal parameter in the environment for the function. This value will be bound to the actual parameter in the environment at the Call site.
input
InOutFun addOne x PushN x
PushI 1
Add
PushN x Bind PushN x Return FunEnd
PushI 1
PushN a
Bind
PushN addOne PushN a
Call PushN a PushI 1 Add Quit
stack
3
2
3 → result of Add (note a is bound to two)
2 → return value of calling addOne and passing in x as an argument → result of binding a
→ result of declaring addOne
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→

6.6 First-Class Functions
This language treats functions like any other value. They can be used as arguments to functions, and can be returned from functions.
6.6.1 Example 1: Curried adder
input
Fun makeAdder x
Fun adder y PushN x PushN y Add
Return FunEnd
PushN adder Return
FunEnd
PushN add3 PushN makeAdder PushI 3
Call
Swap
Bind
PushN add3 PushI 5 Call
Quit
stack
8
→
8 → Evaluated from calling the generated function add3 with argument 5 → The result of binding the generated function to the name add3 → The result of declaring the function makeAdder
Step by step (after declaring makeAdder, pushing add3, pushing 3, and pushing makeAdder):
stack
⟨CLOSURE⟩ add3
stack
add3
⟨CLOSURE⟩
stack
add3
stack

3 makeAdder add3
PushI 5
Call
−−−−→
−−→
stack
stack
Call
Swap Bind PushN add3
−−→
−−−→ −−→ −−−−−−−→
stack
8
5 add3
If a function is returned from another function, it need not be bound to a name in the environment 33

it is returned in. For example:
→
Dunder Mifflin! → Computed from calling the closure returned by the identity function applied to concatExcl with the argument “Dunder Mifflin”.
→ The result of declaring the function _catExcl.
→ The result of declaring the identity function.
Here is a closer look at how the stack develops through this program. Note that function clo- sures will never be on the stack when the program finishes execution.
input
Fun identity x PushN x Return FunEnd
Fun _catExcl y PushS “!” PushN y Concat
Return
FunEnd
PushN identity
PushN _catExcl
Call
PushS “Dunder Mifflin” Call
Quit
stack
Dunder Mifflin!
stack
concatExcl identity
stack
Dunder Mifflin!
⟨CLOSURE⟩
stack
⟨CLOSURE⟩
stack
Dunder Mifflin!
Call PushS “Dunder Mifflin” Call
−−→ −−−−−−−−−−−−−−→ −−→
1. You can make the following assumptions:
• Expressions given in the input file are in correct formats. For example, there will not be expressions like “Push”, “3” or “Add 5” .
• No multiple operators in the same line in the input file. For example, there will not be “Pop Pop Swap”, instead it will be given as
Pop
Pop Swap
• No function closures will be left on the stack.
• All Begin commands will have a matching End.
• There will always be at least one value inside the final stack.
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2. You can assume that all test cases will have a Quit statement at the end to exit your interpreter and output the stack, and that “Quit” will never appear mid-program.
3. You can assume that your interpreter function will only be called ONCE per execution of your program.
Step by step examples
1. If your interpreter reads in expressions from inputFile, states of the stack after each operation are shown below:
input
PushI 10 PushI 15 PushI 30
Sub
PushB Swap
Add Pop Neg Quit
First, Push 10 onto the stack:
Similarly, Push 15 and 30 onto the stack:
Sub will pop the top two values from the stack, calculate 15 – 30 = -15, and Push -15 back:
Then Push the boolean literal onto the stack: 35
stack
10
stack
30 15 10
stack
-15 10

stack
-15
10
Swap consumes the top two values, interchanges them and Pushes them back:
Add will pop the top two values out, which are -15 and , then calculate their sum. Here, is not a numeric value therefore Push both of them back in the same order as well as an error literal
Pop is to remove the top value from the stack, resulting in:
Then after calculating the negation of -15, which is 15, and pushing it back, Quit will terminate the interpreter and write the following values in the stack to outputFile:
Now, go back to the example inputs and outputs given before and make sure you understand how to get those results.
stack
-15 10
stack
-15 10
stack
-15 10
stack
15 10
36

2. More Examples of Bind and Begin…End:
→→
The error is because we are trying to perform an addition on an unbound variable “a”.
input
PushN a PushI 17 Add
stack
17
a
stack
17 a
stack
a
input
Begin PushI 7.2 PushN a1 Bind
End
3.
stack
a1
stack
a1
stack

stack

→→→
input
Begin PushI 3 PushI 7 End PushI 5 Add Quit
4.
stack
7 3
stack
5 7
stack
3
stack
7
stack
12
Explanation :
→→→→
PushI 3 PushI 7
Pushes 3 and 7 on top of the stack. When you encounter the “end”, the last stack frame is saved (which is why the value of 7 is retained on the stack), then 5 is Pushed onto the stack and the values are added.
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7
1.
2.
Frequently Asked Questions
Q: What are the contents of test case X ?
A: We purposefully withhold some test cases to encourage you to write your own test cases and reason about your code. You cannot test every possible input into the program for correctness. We will provide high-level overviews of the test cases, but beyond that we expect you to figure out the functionalities that are not checked with the tests we provide. But you can (and should) run the examples shown in this document! They’re useful on their own, and can act as a springboard to other test cases.
Q: Why does my program run locally but fail on Gradescope? A: Check the following:
• Ensure that your program matches the types and function header defined in section 2 on page 1.
• Make sure that any testing code is either removed or commented out. If your program calls interpreter with input “input.txt”, you will likely throw an exception and get no points.
• Do not submit testing code.
• stdout and stderr streams are not graded. Your program must write to the output file
specified by outputFile for you to receive points.
• Close your input and output files.
• Core and any other external libraries are not available.
• Gradescope only supports 4.04, so any features added after are unsupported.
Q: Why doesn’t Gradescope give useful feedback?
A: Gradescope is strictly a grading tool to tell you how many test cases you passed and your total score. Test and debug your program locally before submitting to Gradescope. The only worthwhile feedback Gradescope gives is whether or not your program compiled properly.
Q: Are there any runtime complexity requirements?
A: Although having a reasonable runtime and space complexity is important, the only official requirement is that your program runs the test suite in less than three minutes.
Q: Is my final score the highest score I received of all my submissions? A: No. Your final score is only your most recent submission.
Q: What can I do if an old submission received a better grade than my most recent submission?
A: You can always download any of your previous submissions. If the deadline is approaching, we suggest resubmitting your highest-scoring submission before Gradescope locks.
3.
4.
5. 6.
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