Out parameters in C

Question

void swap(int &first, int &second){
    int temp = first;
    first = second;
    second = temp;
}

int a=3,b=2;
swap(a,b);

The compiler complaints that void swap(int &first, int &second) has a syntax error. Why? Doesn't C support references?

Answer 1

C doesn't support passing by reference; that's a C++ feature. You'll have to pass pointers instead.

void swap(int *first, int *second){
    int temp = *first;
    *first = *second;
    *second = temp;
}

int a=3,b=2;
swap(&a,&b);

Answer 2

C doesn't support passing by reference. So you will need to use pointers to do what you are trying to achieve:

void swap(int *first, int *second){
    int temp = *first;
    *first = *second;
    *second = temp;
}


int a=3,b=2;
swap(&a,&b);

---

I do NOT recommend this: But I'll add it for completeness.

You can use a macro if your parameters have no side-effects.

#define swap(a,b){   \
    int _temp = (a); \
    (a) = _b;        \
    (b) = _temp;     \
}

Answer 3

for integer swap you can use this method without a local variable:

int swap(int* a, int* b)
{
    *a -= *b;  
    *b += *a;  
    *a = *b - *a; 
}