Let f be the following function: int f(char *s, char *t){char *p1, *p2; for(p1 = s, p2 = t; *p1 ! = ‘\0’& & *p2 ! = ‘\0’; p1++, p2++){if (*p1 == *p2) break; }return p1 –s; }what is the return value of f(“abcd”, “bccd”)?

return value =2.

Here the function f() returns the length of the substring we traversed before we find the same character at the equal index of two substrings.

Take the inputs s= “abcd” and t= “bccd”.  

•Now, p1 points to s1, i.e., p1 points to the character ‘a’ of “abcd”. And similarly, p2 points to ‘b’ of “bccd”.

•Then we compare the values at p1 and p2, are not equal, so p1 and p2 both are incremented by 1.

•Now the characters ‘b’ and ‘c’ of “abcd” and “bccd” respectively are compared. They are not equal. So both p1 and p2 both are incremented by 1.

•Now, p1 points to ‘c’ of “ abcd” that is the element at index 2 of s. And p2 points to ‘c’ of “bccd” that is the element at index 2 of t. Here value at p1 and p2 becomes equal. So the break statement is executed. We stop moving forward.

•As p1 is pointing to index 2 and s is pointing to the base that is index 0, so p1-s = 2.

Explanation:

#include<stdio.h>

int f(char *s, char *t);

void main()

{

int k = f("abcd", "bccd");

printf("%d", k);

}

int f(char *s, char *t)

{

char *p1, *p2;

for(p1 = s, p2 = t; *p1 != '\0'&& *p2 != '\0'; p1++, p2++)

{

 if (*p1 ==*p2)  

 break;  

}

return (p1-s);

}

OUPUT is given as image

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