swap(int ,int ) showing different behavior when calling with and without "&"operator

Question

i am using following code in selection sort program for swap the value.When i pass arguments like swap(a[min],a[j]) It swaps the values but doesn't increment the "swaps" variable. But when i pass arguments like swap(&a[min],&a[j]) It swaps the value as well as increment the "swaps" variable. I didn't get what is happening under the hood.
I am using minGW compiler.

void swap(int *x,int *y)
{
swaps++;
int temp;
temp=*x;
*x=*y;
*y=temp;

}

Yes, actually i was not aware that namespace has it's own swap .Thank you — user3258342, Jun 10 '15 at 06:23
that's [why “using namespace std;” is considered a bad practice](http://stackoverflow.com/q/1452721/995714) — phuclv, Jun 10 '15 at 07:33

score 6 · Answer 1 · answered Jun 10 '15 at 06:17

6

When you use swap(a[min],a[j]), you are most likely calling std::swap. Another reason to avoid

using namespace std;

answered Jun 10 '15 at 06:17

R Sahu

204,454
14
159
270

Now i got it ,Thank you. – user3258342 Jun 10 '15 at 06:22

swap(int *,int *) showing different behavior when calling with and without "&"operator

1 Answers1

swap(int ,int ) showing different behavior when calling with and without "&"operator