Cannot resolve function-like macro inside conditional compilation block

Question

Consider the following - I want to check with #if #endif whether a token is defined somewhere in the code. I am using a CONCAT(input) macro that should glue the constant and changing parts of the token that I want to check.

Unfortunately, the approach presented below causes a compilation error:

error: missing binary operator before token "("

I have found the expressions that can be put inside a #if #endif block:

https://gcc.gnu.org/onlinedocs/cpp/If.html#If

And apparently it states that:

Macros. All macros in the expression are expanded before actual computation of the expression’s value begins.

It turns out that (CONCAT(test)) should be resolved, but it is not.

Is there any workaround allowing to resolve concatenated token names correctly in a conditional compilation block?

#include <stdio.h>

#define CONCAT(input) string##input
#define stringtest 1

int main(void) 
{
    #if defined(CONCAT(test)) && (CONCAT(test)==1)
        printf("OK");
    #else
        printf("NOT");
    #endif

    return 0;
}

I think you can't do that. – unalignedmemoryaccess Jun 08 '17 at 11:31 — unalignedmemoryaccess, Jun 08 '17 at 11:31

Andre Kampling · Accepted Answer · 2017-06-09T07:09:41.670

If you use just: #if CONCAT(test) == 1 it will work and is enought. The statement #if defined(CONCAT(test)) does not work because CONCAT(test) will be evaluated to stringtest which will be evaluated to 1, and you can't use defined on a numerical constant.

A different situation came to my mind - what if I want to check whether the token is eg. != 1 Then if it is not defined anywhere, the condition will evaluate to true. So there would be no difference between token not defined and token being different than 1.

You could handle == 0 equal to not defined. So you could use: #if CONCAT(test) != 0 && CONCAT(test) != 1 where CONCAT(test) != 0 means defined(CONCAT(test)). That is the only alternative, because you can't get macro expansion work in a #ifdef or #if defined() statement, see this question which is very similar to yours.

The gcc documentation says:

Conditionals written like this: #if defined BUFSIZE && BUFSIZE >= 1024

can generally be simplified to just #if BUFSIZE >= 1024, since if BUFSIZE is not defined, it will be interpreted as having the value zero.

Also it can be helpful if you check macro expansion by using gcc -E yourSource.cpp.

gcc --help:

-E Preprocess only; do not compile, assemble or link

Thank you for the simpler suggestion. A different situation came to my mind - what if I want to check whether the token is eg. `!= 1` Then if it is not defined anywhere, the condition will evaluate to true. So there would be no difference between token not defined and token being different than 1. Do you have any suggestion for such a case? — Peter Cerba, Jun 08 '17 at 12:23
I edited my answer to meet your additional question, I hope that helps as an alternative. — Andre Kampling, Jun 09 '17 at 06:08

unalignedmemoryaccess · Answer 2 · 2017-06-08T11:51:10.197

1

You can't do that. Only literals can be used.

You should check for macro directly:

#include <stdio.h>

#define CONCAT(input) string##input
#define stringtest 1

int main(void) {
    #if stringtest  == 1
        printf("OK");
    #else
        printf("NOT");
    #endif
    return 0;
}

Since you don't need defined check, you can use like this:

#define CONCAT(input) string##input
#define stringtest 1

int main(void) {
    #if CONCAT(test) == 1
        printf("OK");
    #else
        printf("NOT");
    #endif
    return 0;
}

edited Jun 08 '17 at 11:51

answered Jun 08 '17 at 11:34

unalignedmemoryaccess

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BTW, this is superfluous. `#if stringtest == 1` is enough. – ArturFH Jun 08 '17 at 11:46

Cannot resolve function-like macro inside conditional compilation block

2 Answers2