How can I pass a string as a preprocessor token?

Question

I am trying to pass a string as a parameter when compiling to this code:

#include <iostream>
#define MY_STRING #STRING_IN

int main(int argc, char** argv) {
    std::cout << MY_STRING;
}

It is compiled with the flag:

-DSTRING_IN=foo

I get the following error:

error: stray '#' in program

 #define MY_STRING #STRING_IN

                   ^

note: in expansion of macro 'MY_STRING'

     std::cout << MY_STRING;

                  ^~~~~~~~~

[...]

Compiler returned: 1

See also here.

How can I pass a parameter via -D... and then turn that into a string?

463035818_is_not_an_ai · Answer 1 · 2018-04-13T06:53:46.937

3

Well, I tried many combinations before posting here, just not the correct one... It works when I stringify the token like this:

#include <iostream>
#define STRINGIFY(x) #x
#define TOSTRING(x) STRINGIFY(x)
#define MY_STRING TOSTRING(STRING_IN)

int main(int argc, char** argv) {
    std::cout << MY_STRING;
}

See also here for the more general case.

edited Apr 13 '18 at 06:53

answered Apr 13 '18 at 06:43

463035818_is_not_an_ai

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There's a redundant layer – Passer By Apr 13 '18 at 06:58
@PasserBy i'd be glad to see a idomatic solution, because this is just the hack i came up with after giving up to really understand what is going on ;) – 463035818_is_not_an_ai Apr 13 '18 at 06:59
1

I meant, you could get rid of the `TOSTRING` layer – Passer By Apr 13 '18 at 07:02
@PasserBy No, that's not redundant. Macro arguments are not expanded when used with `#`. An additional layer is needed if you do want them expanded. – Apr 13 '18 at 07:34
@PasserBy this works, but my trouble started when combining this with passing the token via `-D..` – 463035818_is_not_an_ai Apr 13 '18 at 07:58
@PasserBy Did you actually check the preprocessed result? That evaluates `*"a"`, not `*"b"`. – Apr 13 '18 at 08:21

How can I pass a string as a preprocessor token?

1 Answers1