Printing macro fails to concatenate

Question

I want to create a macro that prints some info and takes any number of arguments to print additional message(s) if needed.

Here's a code snippet im talking about:

#include <stdio.h>

#define print(msg, ...) \
    printf("Line: %d File %s "## msg, __LINE__ , __FILE__, __VA_ARGS__);

int main()
{
    print("Msg: %d", 13);
    print("Msg: %d, Msg2: %d", 123, 234);
}

Here's an error I'm getting:

main.cpp:12:9: error: pasting ""Line: %d File %s "" and ""Msg: %d"" does not give a valid preprocessing token
  printf("Line: %d File %s "## msg, __LINE__ , __FILE__, __VA_ARGS__);

What am I doing wrong?

@paxdiablo the name of the file is main.cpp, so I think it is C style C++ code. — mch, Jul 18 '18 at 11:42
@mch: yes, hence the "pretend" clause in my comment :-) C-style C++ code, as you eloquently put it, is possibly the *worst* thing you can do to learn C++. Having said that, the fact that it is `main.cpp` *does* indicate C++, so I'll restore the C++ tag (and take away the C one) - it's very rare that a question should have both. — paxdiablo, Jul 18 '18 at 11:43

score 2 · Accepted Answer · answered Jul 18 '18 at 11:39

C string literals are joined automatically. The ## you have now is meant to join two "tokens" (not strings) into one, e.g. to take ABC and XYZ and make ABCXYZ as an identifier.

You want this:

printf("Line: %d File %s " msg, __LINE__ , __FILE__, __VA_ARGS__);

score 2 · Answer 2 · answered Jul 18 '18 at 11:40

2

Exactly what it says on the tin: "Line: %d File %s ""Msg: %d" is not a single, valid preprocessing token as ## is required to produce.

Just drop the ##, as adjacent string literals are already concatenated by the preprocessor.

answered Jul 18 '18 at 11:40

Quentin

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Printing macro fails to concatenate

2 Answers2