C++ using "cout" outside of main function?

Question

I don't know if I am just using the wrong keywords.. but I can't find an answer on google. Not can I wrap my feeble mind around my error.

This is a simple demonstration of the error:

#include <iostream>

//std::cout << "hello";

int main()
{
    std::cout << "hello";
    return 0;
}

Upon compiling/running this I receive this error:

main.cpp:3:6: error: 'cout' in namespace 'std' does not name a type

However, if I remove the first cout line, and just allow the program to execute the one inside of the main function, it works just fine.

Anyone got any ideas?

In most programs, code needs to be in a function in order to be executed. — Thomas Matthews, Nov 10 '17 at 22:13

score 5 · Answer 1 · answered Nov 10 '17 at 22:12

5

You cannot run code outside functions in C++. The compiler only expect variable declarations outside functions and, thus, expects std::cout to be a type, which it is not.

answered Nov 10 '17 at 22:12

Jeffrey

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2

"cannot run code" is vague. The syntax does not permit *statements* outside function definitions. – Keith Thompson Nov 10 '17 at 22:16
7

Why of course you can. Just make the code a part of a variable initialisation. `std::ostream& foo = std::cout << "maga";` – n. m. could be an AI Nov 10 '17 at 22:16

C++ using "cout" outside of main function?

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