C Programming - C Preprocessor - Discussion
Discussion Forum : C Preprocessor - Find Output of Program (Q.No. 4)
4.
What will be the output of the program?
#include<stdio.h>
#define JOIN(s1, s2) printf("%s=%s %s=%s \n", #s1, s1, #s2, s2);
int main()
{
char *str1="India";
char *str2="BIX";
JOIN(str1, str2);
return 0;
}
Discussion:
35 comments Page 3 of 4.
Krishz said:
1 decade ago
Ya you are right vasuuma:-).
Prasad Shinde said:
1 decade ago
# means a concatenation operator in C.
It concats 2 strings.
It concats 2 strings.
Nikhil said:
1 decade ago
## concatenates not a single #
Rohit said:
1 decade ago
To concatenate we use dobule hash but single can also do this
Vijendra said:
1 decade ago
What will be the answer the S1, S2?
Wikiok said:
1 decade ago
#define JOIN(s1, s2) printf("%s=%s %s=%s \n", #s1, s1, #s2, s2);
after preprocessing:
printf("%s=%s %s=%s \n", "str1", str1, "str2", str2);
so:
parameter -> parameter
#parameter -> "parameter"
param1##param2 -> param1param2
Or not? :-)
after preprocessing:
printf("%s=%s %s=%s \n", "str1", str1, "str2", str2);
so:
parameter -> parameter
#parameter -> "parameter"
param1##param2 -> param1param2
Or not? :-)
Vishal said:
1 decade ago
I am not getting it. Any one can explain it properly please.
Sachin Kumar said:
1 decade ago
Can anyone explain me that why semi-colon(;) is used because we can't use semi-colon in macros ?
Pooja said:
1 decade ago
Yup.its true..firstly it is concate the 2 string nd than declared with the join s1,s2..
so,#s1 gives output as str1,#s2 gives output as str2
so.str1=india,str2=bix
so,#s1 gives output as str1,#s2 gives output as str2
so.str1=india,str2=bix
Shekar said:
1 decade ago
# operator stringrises the operand. For example #str1 is replaced by "str1" during compilation.
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