118

I am trying to figure out how to write a macro that will pass both a string literal representation of a variable name along with the variable itself into a function.

For example given the following function.

void do_something(string name, int val)
{
   cout << name << ": " << val << endl;
}

I would want to write a macro so I can do this:

int my_val = 5;
CALL_DO_SOMETHING(my_val);

Which would print out: my_val: 5

I tried doing the following:

#define CALL_DO_SOMETHING(VAR) do_something("VAR", VAR);

However, as you might guess, the VAR inside the quotes doesn't get replaced, but is just passed as the string literal "VAR". So I would like to know if there is a way to have the macro argument get turned into a string literal itself.

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1
  • How are you trying to use this? Commented May 8, 2012 at 22:21

5 Answers 5

Reset to default
197

Use the preprocessor # operator:

#define CALL_DO_SOMETHING(VAR) do_something(#VAR, VAR);
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Comments

49

You want to use the stringizing operator:

#define STRING(s) #s

int main()
{
    const char * cstr = STRING(abc); //cstr == "abc"
}
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Comments

15 
#define NAME(x) printf("Hello " #x);
main(){
    NAME(Ian)
}
//will print: Hello Ian
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4 Comments

I’m not totally sure, but it looks like "Hello" #x" (and #x "Hello") causes the string to be glued together without space, which is what is desired in some cases, so this is fairly good answer.
@Smar Make sure you put a space after the constant string Hello: "Hello " #x
Okay I thought so, you should edit that to your answer too since it’s valuable piece of information :)
#define NAME(x) printf("Hello %s", #x);
15

Perhaps you try this solution:

#define QUANTIDISCHI 6
#define QUDI(x) #x
#define QUdi(x) QUDI(x)
. . . 
. . .
unsigned char TheNumber[] = "QUANTIDISCHI = " QUdi(QUANTIDISCHI) "\n";
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4 Comments

How does this answer the question or how is it helpful?
@jirigracik -- It allows to get string presentation of macro expansion as well, unlike other answers
I think it would be useful to explain why having just QUDI(x) is not enough.
This is exactly what I was looking for. The explanation as to why it works would be useful, but it solved my problem (which the other answers did not).
1 
#define THE_LITERAL abcd

#define VALUE(string) #string
#define TO_LITERAL(string) VALUE(string)

std::string abcd = TO_LITERAL(THE_LITERAL); // generates abcd = "abcd"
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2 Comments

Would be nice to get some explanation why we need a two step approach here... not clear to me. But your solution seems to be the only one which works for me :-)
@Klaus, it's to first expand the macro arguments. See here: stackoverflow.com/questions/16989730/… I agree though, this answer could be better explained.

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