Operator to convert to string

Question

Why does the code snippet 1 works but NOT code snippet 2

Code Snippet 1:

#define mkstr(x) #x

int main(void)
{
    printf(mkstr(abc));

    return (0);
}

Code Snippet 2:

int main(void)
{
    printf(#abc);

    return(0);
}

because the # is macro synatx and you don't have a macro in snippet 2 — perivesta, Aug 08 '20 at 08:12
Please stop using macros! We have 2020 and you simply should stop coding like the 70s. — Klaus, Aug 08 '20 at 08:24

score 2 · Answer 1 · answered Aug 08 '20 at 08:12

2

The first snippet works because it has a function-like macro defined in which you put anything, the value is correctly assigned as a constant.

OTOH, the second one has a syntactic error because the compiler doesn't expects that parameter passed in printf(). Thus, the # is meaningless there.

answered Aug 08 '20 at 08:12

Rohan Bari

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You could leave the C++ tag, but I can't argue with your edit, C and C++ are different languages and the separation is encouraged, like @klutt stated int the accepted answer. – anastaciu Aug 08 '20 at 09:42
@anastaciu the question is explicitly asked for C. That's why I had removed C++ tag and nothing. If you see something irrelevant in the edit, then you can take it back. No problem at all. – Rohan Bari Aug 08 '20 at 10:14
1

No, I think it's ok. – anastaciu Aug 08 '20 at 10:15

Debargha Roy · Accepted Answer · 2020-08-08T12:25:51.897

Commands starting with the # symbol are called Macros in C and C++. Macros are blocks of code that are named and referenced with that name.

There are 2 popular types of macros - the Object-like and the function-like. The one you're using is the function-like macro.

The Preprocessor is responsible for replacing macro calls with the actual object/function calls.

The statement in Snippet 1

#define mkstr(x) #x

The above macro uses a special feature called the stringizing. The # before the x specifies that the input parameter should be treated as is, ie. converted to a string constant, thereby returning a string equivalent of what is passed.

On the contrary, when you call the below code in Snippet 2

printf(#abc);

doesn't mean anything. It's a compiler error as #s are not allowed in the middle or end of a statement (please not that # is allowed to be part of string such as "#1", or when used as a character literal as '#'). And thus any statement that starts with # becomes a macro.

Caution: Use of macros is discouraged. You can refer this answer on StackOverflow on why not to use them.

Refer the below resources to learn more about macros

Operator to convert to string

2 Answers2