why swapping value using pointer doesn't work?

Question

I have written the following code. It should swap the values of two variables ... but, as soon as I compile the code it shows swap.exe has stopped working...
Why doesn't it work?

#include<cstdio>
#include<iostream>
using namespace std;
void swap(int *x, int *y)
{
    int *temp;
    *temp=*x;
    *x=*y;
    *y=*temp;
}
int main()
{
    int i=5,j=10;
    swap(&i,&j);
    cout<<i<<" "<<j<<endl;

    return 0;
}

How to fix this problem? What's wrong with my code?

Answer 1

Change to

void swap(int *x, int *y)
{
    int temp;
    temp=*x;
    *x=*y;
    *y=temp;
}

If you want to swap two int s why do you use a pointer to int as temporary variable? Use the same type.

Answer 2

Because temp is a dangling pointer, so accessing *temp is undefined behavior.

You can use int temp; and temp = *x; ... temp = *x; ... , or just use std::swap .

Answer 3

It does not work, because you copy the content of x to a location that is pointed by an uninitialized pointer ( temp ).

If you simply want to swap the contents of x and y you may use std::swap(*x, *y) .

Answer 4

use temp as int not as int * as in:

void swap(int *x, int *y)
{
    int temp;
    temp=*x;
    *x=*y;
    *y=temp;
}

Answer 5

declare int temp as just a int type variable, not a pointer. Because you are just storing value of x in temp, not pointing to x.