How to use Macro argument as string literal?

Question

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.

How are you trying to use this? – chris May 08 '12 at 22:21 — chris, May 08 '12 at 22:21

Morwenn · Accepted Answer · 2015-05-12T14:08:09.213

183

Use the preprocessor # operator:

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

edited May 12 '15 at 14:08

answered May 08 '12 at 22:12

Morwenn

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score 45 · Answer 2 · answered May 08 '12 at 22:12

45

You want to use the stringizing operator:

#define STRING(s) #s

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

answered May 08 '12 at 22:12

chris

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score 15 · Answer 3 · answered Aug 23 '16 at 18:32

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";

answered Aug 23 '16 at 18:32

Zili

197
1
2

How does this answer the question or how is it helpful? – jirig Apr 02 '18 at 19:21
2

@jirigracik -- It allows to get string presentation of macro expansion as well, unlike other answers – grepcake Jun 15 '18 at 18:06
10

I think it would be useful to explain why having just `QUDI(x)` is not enough. – LRDPRDX Feb 06 '20 at 07:55
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). – GMc Dec 02 '20 at 08:16

score 14 · Answer 4 · answered Oct 01 '14 at 22:31

14

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

answered Oct 01 '14 at 22:31

Mikele Shtembari

163
2
7

1

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 Jun 09 '17 at 13:37
2

@Smar Make sure you put a space after the constant string Hello: `"Hello " #x` – jack Jun 11 '17 at 00:37
Okay I thought so, you should edit that to your answer too since it’s valuable piece of information :) – Smar Jun 11 '17 at 06:00
#define NAME(x) printf("Hello %s", #x); – Liam de Koster-Kjaer Dec 01 '21 at 09:11

score 0 · Answer 5 · answered Mar 27 '23 at 10:51

0

#define THE_LITERAL abcd

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

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

answered Mar 27 '23 at 10:51

Audrius Meškauskas

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How to use Macro argument as string literal?

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