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;
}
str1=IndiaBIX str2=BIX
str1=India str2=BIX
str1=India str2=IndiaBIX
Error: in macro substitution
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
35 comments Page 4 of 4.

Kundan said:   1 decade ago
I agree with Natasa. For better understanding purpose I give a example code.

#include<stdio.h>
#define res(s1,s2) printf("%s=%d %s=%d",#s1,s1,#s2,s2);

void main()
{
int a=4,b=3;
res(a,b);
}

Output of this code is
a=4 b=3

Saurabh Tiwari said:   1 decade ago
Yup.... arun and puja both of you are right because

#define JOIN(s1, s2) printf("%s=%s %s=%s \n", #s1, s1, #s2, s2);

Here

1) #s1 is equal to "variable name" i.e (str1) so it will print same as it is.

2) s1 that will print the "value of str1" i.e (India along with =)
and so on.

Vimala said:   1 decade ago
# defines that name of the variable.

Durga dutt said:   1 decade ago
isn't an error if we use ";" at the end of macro expansion?

Nachiyappan said:   2 decades ago
Is it possible to print value through pre-processor ?


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