A function call is made in the code. In the first function call a pass by reference is performed calling the pointer function. In the second function call a pass by value is performed where the reference function is called.
Why is this so?
#include <iostream>
void f(int *p) { (*p) ; }
void f(int &p) { p-=10; }
int main() {
int x=0; f(&x); f(x); f(&x);
std::cout << x << "\n";
}
CodePudding user response:
x is an int variable. &x is taking the address of x, yielding an int* pointer.
f(&x) can't pass an int* pointer to an int& reference, but it can pass to an int* pointer, so it calls:
void f(int*)
f(x) can't pass an int variable to an int* pointer, but it can pass to an int& reference, so it calls:
void f(int&)
