swap(int &a, int &b) and swap(int *a, int *b). What is difference?

Question

void swap(int *a, int *b)
{
    int iTemp ;
    iTemp = *a;
    *a = *b;
    *b = iTemp;

}
void swap2(int &a, int &b)
{
    int iTemp;
    iTemp = a;
    a = b;
    b = iTemp;
}

What are the difference between swap2(&a, &b) and swap(*a,*b). Although, final result is the same.

Answer 1

void swap(int *a, int *b)

You are expecting the variable to pass by pointer (a value of the type pointer to int)

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void swap2(int &a, int &b)

you are expecting the variable to pass by reference (which is the the address of an integer variable)

---

However, the two previous statements are the same technically.

1) The first one is called like this:

int a,b;
swap(&x,&y);

the other one:

int a,b;
swap2(x,y);

2) You can pass A NULL or nullptr to the first one. However, you can not for the second one.

As far as I remember, in google c++ style guide, they prefer the first one since it is obvious that the parameter may be changed.