Print a #define macro using std::cout

March 30, 2022

I am trying to do this

#define _TEST_ test
#include <iostream>

int main()
{
        std::cout << "_TEST_" << std::endl;
}

As far as my understanding, I expect this output.

test

However, the output I get is

_TEST_

Why am I doing wrong here?

>Solution :

"_TEST_" is a string literal and not a macro. So no macro replacement will be done due to "_TEST_". To achieve your expected output you need to remove the surrounding double quotes and also change the macro to as shown below

//-------------vvvvvv--->double quotes added here
#define _TEST_ "test" 
#include <iostream>

int main()
{
//-------------------vvvvvv---------------> not withing quotes
        std::cout << _TEST_ << std::endl;
}

The output of the above modified program is:

test

Demo

Explanation

In the modified program, the macro _TEST_ stands for the string literal "test". And thus when we use that macro in the statement std::cout << _TEST_ << std::endl;, it will be replaced by the string literal, producing the expected output.