depth in std::cout statement in C++

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Hi,

#include<iostream.h>

using namespace std;

int main()
{

std::cout<<"test application:"<<std::endl;

return 0;
}

How std::cout and std::endl is working internally. How "test application" is printing in console application.

Regards,
Ranjith

Posted 27-Oct-13 20:47pm

ranjithkumar81

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Andreas Gieriet · Answer 1 · 2013-10-27T20:57:00

Solution 1

Have a look into the include files.
cout finally maps into STDOUT, which is one of the io streams that maps to the console by default. It's a system call that takes care of this mapping.
Cheers
Andi

Posted 27-Oct-13 20:57pm

Andreas Gieriet

Comments

Sergey Alexandrovich Kryukov 28-Oct-13 3:22am

My 5, but OP would probably needs explanation of how the operator is defined...
—SA

Richard MacCutchan · Answer 2 · 2013-10-27T23:13:00

Solution 2

See http://msdn.microsoft.com/en-us/library/h07e2k12.aspx[^].

Posted 27-Oct-13 23:13pm

Richard MacCutchan