Passing variables by pointers...what am I doing wrong?

[Shadow12345]

OK I am simply trying to pass variables by pointers but for some reason it doesn't work, can someone tell me what to change?

#include <iostream>
using namespace std;

void Swap(int*, int*);
int main(void) {
	int x, int y, int swapped;
	cout << "Enter two variables" << endl;
	cin >> x >> y;
	cout << "You have entered " << x << " and " << y << "!";
	Swap(&x,&y);
	cout << "X is now " << x << " Y is now " << y << endl;
	return 0;
}

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

Thanks if you can help me...it is really important that I can be able to EFFECTIVELY pass variables using pointers. When you get into large programs simply passing the variables themselves can slow the program down.

[1943]

Hmm that looks familiar to a problem in the book I'm working in..
(Teach Yourself C++, 21days)

I've modified it to this...hope that helps

#include <iostream>
using namespace std;

void Swap(int*, int*);

int main(void) {
	// initialize variables (gets rid of warnings)
	int x = 0, y = 0;

	cout << "Enter two variables" << endl;
	cin >> x >> y;
	cout << "You have entered " << x << " and " << y << "!" << endl;
	Swap(&x,&y);
	cout << "X is now " << x << " Y is now " << y << endl;
	return 0;
}

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

I'm not that advanced, so I can't tell you why it doesn't work the other way...

[Magos]

int x, int y, int swapped;
make this to int x, y, swapped;

Your real problem lies in the swap function. You define a pointer temp, but you never tell it what to point at. And then you try to set it to a value. This means you're trying to set a value at a random place in memory, which will (most likely) cause a crash.

Make this pointer a normal int instead, and it will work.

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