How to get the definition of a macro as a string literal? - c++

Question

Say in a header, which I do not want to read myself but which I do include, I have
#define A B
#define B C
Now
#define STR(name) # name
defines a macro that gives me the name of any macro as a string, and
#define EXP_STR(name) STR(name)
defines a macro that gives me the full expansion of any macro as a string. So
cout << STR(A) << EXP_STR(A) << endl;
will print AC.
Is there any way to get "B" from A using some macros?

Answer 1

Since you can write
#define B C
#define A B
#define STR(name) # name
#define EXP_STR(name) STR(name)
and the
cout << STR(A) << EXP_STR(A) << endl;
will output exaclty the same, it means that it's not possible.
When you do this
#define A B
and then
#define B C
now this means that A will be substituted by C and not B, so there will be no way to do it because when the cout line is reached the preprocessor had already substituted A by C.
So the short answer is, it's not possible because the preprocessor would have replaced A with C before the file is compiled.

Answer 2

Yeah, it's possible. You just have to use a couple of cheats.
#undef B before #define EXP_STR
Use a few more levels of indirection
For example:
#define A B
#define B C
#define _TEMP_VAR B // so that it can be redefined later
#undef B // so that EXP_STR(A) will return "B"
#define EXP_STR__(x) (x)
#define EXP_STR_(x) EXP_STR__(#x)
#define EXP_STR(x) EXP_STR_(x)
#define STR(x) # x
#define B _TEMP_VAR // now you can still access A normally, defined to B (defined to C)
A test program to prove it:
#include <stdio.h>
int main(void)
{
printf( "EXP_STR(A) = %s\n", EXP_STR(A) );
printf( "STR(A) = %s\n", STR(A) );
}
Output:
EXP_STR(A) = B
STR(A) = A