CODEWITHSUNDEEP
cc-preprocessorstringification
Vijay
Vijay

Reputation: 67291

What does ## mean for the C(C++) preprocessor?

I have a C program below:

#define f(g,g2) g##g2
main()
{
int var12=100;
printf("%d",f(var,12));
}

when I run just the preprocessor it expands this as 

{
int var12=100;
printf("%d",var12);
}

which is the reason why the output is 100.

Can anybody tell me how/why the preprocessor expands var##12 to var12?

Upvotes: 17

Views: 25767

Answers (4)

cobbal
cobbal

Reputation: 70775

Nothing too fancy: ## tells the preprocessor to concatenate the left and right sides.

See here.

Upvotes: 37

SHANAVAS P
SHANAVAS P

Reputation: 31

#define f(g,g2) g##g2

## is usued to concatenate two macros in c-preprocessor. So before compiling f(var,12) should replace by the preprocessor with var12 and hence you got the output.

Upvotes: 2

Stefano Borini
Stefano Borini

Reputation: 143925

because ## is a token concatenation operator for the c preprocessor.

Or maybe I don't understand the question.

Upvotes: 6

Prasoon Saurav
Prasoon Saurav

Reputation: 92924

## is Token Pasting Operator

The double-number-sign or "token-pasting" operator (##), which is sometimes called the "merging" operator, is used in both object-like and function-like macros. It permits separate tokens to be joined into a single token and therefore cannot be the first or last token in the macro definition.

If a formal parameter in a macro definition is preceded or followed by the token-pasting operator, the formal parameter is immediately replaced by the unexpanded actual argument. Macro expansion is not performed on the argument prior to replacement.

Upvotes: 6

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