Why does the C preprocessor consider enum values as equal?

Question

Why does the std::cout line in the following code run even though A and B are different?

#include 

en         


        

Answer 1

This is because the preprocessor works before compile time.

As the enum definitions occur at compile time, A and B will both be defined as empty (pp-number 0) - and thus equal - at pre-processing time, and thus the output statement is included in the compiled code.

When you use #define they are defined differently at pre-processing time and thus the statement evaluates to false.

In relation to your comment about what you want to do, you don't need to use pre-processor #if to do this. You can just use the standard if as both MODE and MODE_GREY (or MODE_RGB or MODE_CMYK) are all still defined:

#include 

enum T { MODE_RGB = 1, MODE_GREY = 2, MODE_CMYK = 3 };

#define MODE MODE_GREY

int main()
{
    if( MODE == MODE_GREY )
        std::cout << "Grey mode" << std::endl;
    else if( MODE == MODE_RGB )
        std::cout << "RGB mode" << std::endl;
    else if( MODE == MODE_CMYK )
        std::cout << "CMYK mode" << std::endl;

    return 0;
}

The other option using only the pre-processor is to do this as @TripeHound correctly answered below.