[SOLVED] CS计算机代考程序代写 // topics: references, reference parameters, operator precedence.

// topics: references, reference parameters, operator precedence.

#include
using namespace std;

int f1(int f1p) {
// Here, a copy of the passed value is made upon call.
// The value manipulated is a local value.
// Changing the local value does not affect the passed value.
f1p = f1p+1;
// A copy of the returned value is made upon return.
return f1p;
}
int& f2(int& f2p) {
// Here, the parameter is a reference to the passed value.
// The value manipulated is the same as the value that was passed.
f2p = f2p+2;
// The reference is returned, so that it can be used as a value in an expression.
return f2p;
}

int main() {
int referredto = 0;
// after this, x is an alias to referredto.
// If we change the value of x, we change the value of referredto, and inversely.
int& x = referredto; ;
//
int y = 1;
cout << “x, referredto, y : ” << x << referredto << y << endl; // Here, it may see that we are now making x referring to y instead of referredto. // No. We are assigning the value of y to x, which is still an alias to referredto.// x is now equal to 1, so is referred to. x = y; cout << “x, referredto, y : ” << x << referredto << y << endl;// If we change the value of x, we dont change the value of y;x = 0; cout << “x, referredto, y : ” << x << referredto << y << endl;cout << “f1(x): ” << x << f1(x) << x << endl;x = 0; cout << “f1(x+1): ” << x << f1(x + 1) << x << endl; x = 0; // Here, it seems x was changed prior to the call to f2. // What really happens is that when the whole expression evaluates, // the function call to f2 has the highest priority, so it is called // before any of the << operators. Thus, the value of x is changed // by f2 (as it is passed by reference) before the initial value of x// is printed. cout << “f2(x): ” << x << f2(x) << x << endl; x = 0;// one way to print what we wanted, we need to separate the statement. cout << “f2(x): ” << x;cout << f2(x);cout << x << endl;x = 0;// Here we show that the consecutive stream insertion operators// are evaluated from right to left if they involve expressions that// have operations with the same precedence. cout << “f2(x): ” << f2(x) << f2(x) << f2(x) << endl;x = 0;// ++x has higher precedence than <<, so it takes effect before them. // incrementing x before returning the incremented x, then all the <