CODEWITHSUNDEEP
c
Tim
Tim

Reputation: 99418

Nested macros and ##

From The C Programming Language, by KRC

After 

#define cat(x, y)       x ## y

the call cat(var, 123) yields var123. However, the call cat(cat(1,2),3) is undefined: the presence of ## prevents the arguments of the outer call from being expanded. Thus it produces the token string cat ( 1 , 2 )3 and )3 (the catenation of the last token of the first argument with the first token of the second) is not a legal token.

If a second level of macro definition is introduced, 

#define xcat(x, y)      cat(x,y)

things work more smoothly; xcat(xcat(1, 2), 3) does produce 123, because the expansion of xcat itself does not involve the ## operator.

What is the property of ## that makes the difference between the two examples?

Why is the inner cat(1,2) in the first example not expanded, while the inner xcat(1,2) in the second example is?

Thanks!

Upvotes: 3

Views: 636

Answers (1)

tofro
tofro

Reputation: 6063

It is one of the (not-so-well-known) characteristics of the macro ## operator that it inhibits further expansion of its arguments (it just considers them plain strings). An excerpt from the gcc pre-processor docs:

...As with stringification, the actual argument is not macro-expanded first...

That is, arguments to ## are not expanded.

By implementing the additional indirection using your xcat macro you are working around the problem (A process that is called the argument prescan is jumping in and actually evaluates the resulting string twice)

Upvotes: 1

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