Pass Reference C

Transferred from: http://myturn.blog.hexun.com/15584978_d.html

#include <iostream>

using namespace Std;

void Swap (int x, int y);

int main (void)

{

int a = 1;

int b = 2;

cout << "a =" << a << "," << "b =" << b << Endl;

Swap (A, b);

cout << "a =" << a << "," << "b =" << b << Endl;

System ("pause");

return 0;

}

One: Value passing

void Swap (int x, int y)

{

int temp = x;

x = y;

y = temp;

}

Output Result:

A = 1, b = 2

A = 1, b = 2

Cause: the Swap (int x, int y) function takes a value pass, and the actual arguments passed in are actually copies of a and b rather than themselves, so changes to the copy do not reflect A and B.

II: Reference Delivery

void Swap (int &x, int &y)

{

int temp = x;

x = y;

y = temp;

}

Output Result:

A = 1, b = 2

A = 2, B = 1

Cause: the Swap (int x, int y) function takes the form of a reference pass, and the actual arguments passed in are actually references to A and B, and the changes to the reference directly reflect A and B.

Three: pointer passing

1. Change the pointer itself

void Swap (int *x, int *y)

{

int *temp = x;

x = y;

y = temp;

}

Calling method: Swap (&a, &b);

Output Result:

A = 1, b = 2

A = 1, b = 2

Cause: the Swap (int x, int y) function takes a pointer pass, and the actual argument passed in is actually a copy of the pointer to A and B, and it changes the copy itself rather than its indirect reference, so it does not affect the value that the pointer points to, A and B.

2. Changing the pointer's indirect reference

void Swap (int *x, int *y)

{

int temp = *x;

*x = *y;

*y = temp;

}

Calling method: Swap (&a, &b);

Output Result:

A = 1, b = 2

A = 2, B = 1

Cause: the Swap (int x, int y) function takes a pointer pass, although the incoming argument is also a copy of the pointer to a and B, but changes the indirect reference to the copy, either the pointer itself or its copy, pointing to the same value, so the change will reflect A and B.

Pass reference C (GO)