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 <
