[Solved] C pointer—function relation [duplicate]

Question

The difference is the type of the parameters you call the function with.

First case:

The type of x is int and the foo function is expecting int *.

So you have foo(&x, &y);

Second case:

The type of p1 is int * and the swap function is also expecting int *.

So you have swap(p1, p2);

Accepted Answer

The difference is the type of the parameters you call the function with.

First case:

The type of x is int and the foo function is expecting int *.

So you have foo(&x, &y);

Second case:

The type of p1 is int * and the swap function is also expecting int *.

So you have swap(p1, p2);