Simple C++ swap function

Question

Why is it that if I have a function like this, to swap two numbers, it doesn't work[swap], (I know I can do this by declaring pointers in the prototype, and then pass the address of the respective variables in main()), but works for array, without having to pass pointers and addresses.

Doesn't work

void num_exchange(int m, int n);

int main(){

int num1 = 5;
int num2 = 6;

num_exchange(num1 , num2 );

cout << "num1 =" << num1 << endl;
cout << "num2 =" << num2 << endl;

return 0;
}

void num_exchange(int m, int n){
int temp;
temp = m;
m = n;
n = temp;
}

Works

void arr_exchange(int [], int);

int main(){


int n[7] = { 0, 0, 0, 0, 0, 0, 0 };

arr_exchange(n, 7);
for (int i = 0; i < 7; i++)
    cout << n[i] << " ";

return 0;
}

void arr_exchange(int x[], int){
for (int i = 0; i < 7; i++)
    x[i] = 1;
}

Answer 1

void num_exchange(int m, int n){
int temp;
temp = m;
m = n;
n = temp;
}

modifies copies of the input integers. To make your code work use

void num_exchange(int& m, int& n){
int temp;
temp = m;
m = n;
n = temp;
}

instead (note the & in the first line). This is called passing by reference. In general, use std::swap to swap things.

void arr_exchange(int x[], int){
for (int i = 0; i < 7; i++)
    x[i] = 1;
}

works because in C++

void arr_exchange(int x[], int){

is equivalent to

void arr_exchange(int* x, int){

So here a pointer is passed and thus the original data is modified.