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 3 of 4.

Gaurav Kumar Garg said:   1 decade ago
@Sachin Kumar.

Its your choice you can also remove semicolon. without semi colon it will also work.

With semi colon in macro
You can also call JOIN(str1, str2) <--- no semicolon.

Avinash said:   1 decade ago
# symbol is used to print the actual arguments that are passed to function.

Sunil Kothari said:   1 decade ago
#include<stdio.h>
#define cube(x) (x*x*x)

void main()

{
int a,b =3;
a= cube(++b);
printf("%d \n %d", a, b);
}

Output:
-------
150
6

Please explain above program?

Shashwat said:   1 decade ago
We have to add semicolon either at the macro declaration or while calling macro.

Neeraja said:   1 decade ago
Please give clarity what #s1 #s2 means?

Omkar said:   1 decade ago
Can anyone explain @Sunil's program ?

Bhushan said:   1 decade ago
@Sunil : It should be:

216.
6.

Value would be overwrite.

a = (++b * ++b * ++b).
a = (4 * ++b * ++b).

a = (5 * 5 * ++b).
a = (6 * 6 * 6).
a = 216.

Nimisha said:   1 decade ago
#s is used to print string value.

Kaviyakarthi said:   10 years ago
@Sunil:

I thought the solution is,

a = ++b*++b*++b.
a = ++3*++b*++b.
a = 4*++4*++b.
a = 4*5*++5.
a = 4*5*6.
a = 120.
b = 6.

Kaviyakarthi said:   10 years ago
@Sunil:

I thought the solution is,

a = ++b*++b*++b.
a = ++3*++b*++b.
a = 4*++4*++b.
a = 4*5*++5.
a = 4*5*6.
a = 120.
b = 6.


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