Why does this output "geeksforgeeks"?

Question

#include <iostream>
using namespace std;
int main()
{
    if (!(cout << "geeks"))
       cout <<" geeks ";
    else
       cout << "forgeeks ";

    return 0;
}

Why is cout << "geeks"; inside the if condition executed? I know that the if statement is false. I expected "forgeeks " only.

The expression `(cout << "geeks")` in the `if` clause has side effects - it prints "geeks". — Max Langhof, May 09 '18 at 10:58
To put it another way: `(cout << "geeks")` in the `if` is executed to determine the outcome and that `cout` prints as well. — Stefan, May 09 '18 at 11:00
Why do you *not* expect `if (!(cout << "geeks"))` to be executed? — David, May 09 '18 at 11:04
Your time is better spent with [these C++ books](https://stackoverflow.com/questions/388242/the-definitive-c-book-guide-and-list). — Ron, May 09 '18 at 11:21

score 5 · Accepted Answer · answered May 09 '18 at 11:07

Why is cout << "geeks"; inside the if condition executed?

Because otherwise the computer won't know whether it was "true" or "false"?

Given if (foo()), the function foo must be called; this extends to any expression in general, which must be evaluated before their "result" can be known (though note that sub-expressions may be skipped due to short-circuiting).

score 0 · Answer 2 · answered May 09 '18 at 11:19

Focus in below statement:

if (!(cout << "geeks"))

Here cout with << operator which is overloaded to print the stream as output i.e "geeks" and then it return this stream to the if statement.

if statement check the condition i.e if(!("geeks")), which if statement see as if(!(true)), results to false condition.

Hence else statement execute to print "forgeeks".

Why does this output "geeksforgeeks"?

2 Answers2