C//swap does not work

I realized that my function was just swapping parameters without affecting the arguments.

Why do I need to add the address character?

Changing the parameter does not change the argument
If you need to change the argument, the argument type should be a pointer

All arguments to the C function are passed as a "value call", which means that the function gets a copy of the parameter values. If you need to change the value of an argument, the argument type should be a pointer. In this case, you need to add an address.

Line 4 with & before x and y;
because it does not use the reference symbol & What is changed is only the substitute of x and y, not the change itself. Plus & If you don't use a surrogate, changes inside the function affect itself.