CareerCup

12
concatinate(1,2)

wth kind of question is this?

Man I've been a developer for 10 years and I've never run into a macro with # in it. I've been on large code bases too.

I wonder why amazon even bothers with this question? If the candidate gets it wrong, what does that mean? If he/she gets it right, what does that mean?

Really amazon, get a clue here.. :)

I had the exact same feelings as above person when I saw this question.
It doesn't tell much at all about the interviewee. Even if s/he gets it wrong:
1) He may just got confused
2) He may just didn't know ## crap
3) He hates macros anyways, so => Make him the director of a S/w org at Amazon LOL..

# is used for stringifying
## is used for concatenation

Ordinary arguments to macros are COMPLETELY expanded before stringification is applied.

answer is 12 in both the cases........thnx

#a means "a"
a##b means ab thus #(1##2)=========>>"12"

concatinate(1,2)
concatinate(1,2)

$ cat test2.c

# include <stdio.h>

#define concatinate(a,b) a##b
#define same1(a) #a
#define same2(a) same1(a)

int main(){
printf("%s\n",same2(concatinate(1,2)));
printf("%s\n",same1(concatinate(1,2)));
}

$ cc test2.c
$ ./a.out

12
concatinate(1,2)

When the compiler comes to this line:
printf("%s\n",same2(concatinate(1,2)));
it sees same2 for which it replaces it with same1.

Now, it sees concatinate(1, 2). It immediately replaces it with 12.

When it comes to same1, it sees a definition and replaces the rest of the code with #a i.e. concatinate(1, 2).

So, the compiler actually does the same2 in pass two. And same1 in pass one.