GATE Overflow Test Series | Programming | Test 1 | Question: 4

Question

The number of characters (including whitespaces if any) printed by the following C program is

#define sum(a, b) #a "+"#b "=%d"
#include<stdio.h>
int main()
{
  printf(sum(6,9), 6+9);
}

 

Answer 1

C preprocessor automatically concatenates adjacent strings. '#' operator makes a string and so sum(6,9) expands to "6" "+" "9" "=%d" which concatenates to "6+9=%d". So, the final printed value will be "6+9=15". 6 characters.

Comments

Learned these 2 points in this question : 

1.) “sdfabc” = “sdf” “abc” in printf() only.

2.) # add’s “” around a name in macro’s only.

---

printf(“sdf” “abc”)  = sdfabc, this is not equal to printf(“sdf” + “abc”);

refer the images.

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@Udhay_Brahmi it is not only in printf(), in C Adjacent string literal tokens are concatenated.

#include<stdio.h>
int main()
{
  char str[] = "Hello""hi";
  char *p=str;
  while(*p != '\0')
  {
   printf("%c",*p);
   p++;
  }
 }

If we run this code o/p will be Hellohi

Answer 2

printf(sum(6,9), 6+9); // it will be converted to printf(“6+9=%d”,6+9 ); // “6+9=15”