C complex #define question

[pronesti]

What does this statement do? I know ## concatenates and I know the converts macro parameters to string literals, but overall I'm not sure what this statement does. Any help would be appreciatede.



#define ADD_TOKEN_LIST(name) { "Xig" #name "token", Xig ## name ## tokens }

[NUj6URfi]

Defines a statement.

[pronesti]

I know it defines a statement, but I don't understand the statement.

[Disch]

Not very helpful, toad... >_>


Anyway... to answer your question.

#name takes the parameter name and turns it to a string literal.
## takes two separate tokens and combines them into one.

For example... if you were to say:

ADD_TOKEN_LIST(foo)

That would expand to the below:

{ "Xig" "foo" "token", Xigfootokens }
          ^                 ^
          |                 |
          |             ## name ##:  name gets replaced with foo
          |             then the ##s make the foo 'join' with the tokens around it
          |
          |
          #name takes the given name 'foo' and turns it into a string literal 

Also... in C and C++... two string literals that are declared next to each other like that are automatically joined. This is true everywhere, not just in macros.

So "Xig" "foo" "token" essentially becomes "Xigfootoken"


Therefore....

ADD_TOKEN_LIST(foo)

// expands to...

{ "Xigfootoken", Xigfootokens }

EDIT:

Usage of this is likely to build an array. IE, it's probably used like this:

struct SomeStruct[] = 
{
ADD_TOKEN_LIST(foo),
ADD_TOKEN_LIST(bar),
ADD_TOKEN_LIST(baz)
};

[pronesti]

Thanks, and have a great day!