CODEWITHSUNDEEP
c++
Trishal Mandrik
Trishal Mandrik

Reputation: 1

Why cout statement is not executing in passing pointer to function?

The cout statement in function call swap is not executing. Why?

#include <iostream>

using namespace std;

void swap(int *a, int *b) {
  int temp = *a;
  *a = *b;
  *b = temp;
  cout << "Value Swapped" << endl;
}

int main() {
  int x = 400, y = 100;
  swap(x, y);
  cout << "x = " << x << " y = " << y << endl;
  return 0;
}

I have expected below output:

Value Swapped
x = 100 y = 400

But the output I have got is: 

x = 100 y = 400

Upvotes: 0

Views: 194

Answers (4)

Aconcagua
Aconcagua

Reputation: 25516

using namespace std imports another equally named function from namespace std into global namespace, so you end up in having two overloads:

swap(int&, int&); // from std; actually a template instantiation
swap(int*, int*);

You call swap as

swap(x, y);

Now consider: Which are the types of x and y, are they pointers? So decide yourself, which overload will get called?

If you hadn't imported the second overload via using namespace std, which generally is considered bad practice anyway (well, you got a victim of the pitfalls of right away...), you would have ended up in a compilation error instead (integers aren't converted to pointers implicitly!).

To get your function called, you need to make pointers from:

swap(&x, &y);
//   ^   ^

Actually, that would have worked even with the imported namespace (despite of being bad practice), as the imported overload is a template, yours an ordinary function, thus yours is considered the more specialised overload and thus will be preferred.

Upvotes: 0

C. R. Ward
C. R. Ward

Reputation: 93

Consider choosing a unique enough identifier name such that there will not be a conflict with a namespace identifier (as in std::swap). If you can't choose the name of your identifier, put it inside of a namespace.

Secondly, if the compiler does not enforce the function declaration that you state, you should manually check to see that you passed the variables correctly (i.e., &x and &y instead of x and y).

Upvotes: 0

cHao
cHao

Reputation: 86506

You're not seeing "Value swapped" because your swap function never gets called. Instead, you're calling std::swap<int>.

If you hadn't said using namespace std;, you'd have found out that your swap function is being passed integers, but was expecting pointers. Reason #1 not to use using namespace std; -- you can easily end up not calling what you think you are. :)

Upvotes: 3

Sudip Sarker
Sudip Sarker

Reputation: 421

your function swap is expecting pointers as arguments, but you are passing integers. So instead of calling your function, it's calling the c++ swap function. To better understand this, change the function name 'swap' to any other name like 'swapp' and you'll see this error: 

invalid conversion from int to int*

write swap(&x, &y) and you'll get your desired output

Upvotes: 0

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