#define cat(x,y) x##y concatenates x to y. But cat(cat(1,2)...

FAQS.ORG makes no guarantees as to the accuracy of the posts. Each post is the personal opinion of the poster. These posts are not intended to substitute for medical, tax, legal, investment, accounting, or other professional advice. FAQS.ORG does not endorse any opinion or any product or service mentioned mentioned in these posts.

<< Back to general questions

[ Home | FAQ-Related Q&As | General Q&As | Answered Questions ]

Question by new_2_c
Submitted on 9/15/2003
Related FAQ: N/A
Rating: Not yet rated	Rate this question:
#define cat(x,y) x##y concatenates x to y. But cat(cat(1,2),3) does not expand but gives preprocessor warning. Why?

Answer by wbic16
Submitted on 10/31/2003
Rating: Not yet rated	Rate this answer:
You are using a macro, which was never intended to be used as a function. Define a function and use that instead. The preprocessor traverses your code and replaces instances of cat(x,y) with x##y. I'm guessing that your preprocessor is doing this only once and spitting out cat(1,2)##3 instead of 1##2##3 as you intended...

Your name or nickname:
If you'd like to create a new account or access your existing account, put in your password here:
Your answer: