GATE CSE 2017 Set 2 | Question: 55

Question

Consider the following C program.

#include<stdio.h>
#include<string.h>
int main() {
    char* c="GATECSIT2017";
    char* p=c;
    printf("%d", (int)strlen(c+2[p]-6[p]-1));
    return 0;
}

The output of the program is _______

Comments

Please note here that you don’t really need to remember the ASCII values of T or I. Why?

Because you just have to know the number of alphabetical difference between the two characters T & I. The difference is simply 11 (you can even count it on your fingers). Assuming A → 1; B → 2; C → 3;….. Z → 26

I.e; what’s the difference between G & D ?
‘G’ – ‘D’:   7 – 4 = 3

This intuition would come in handy in the exam environment.

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Key concepts to note:

1. 2[p] = *[p+2] = p[2]
2. 6[p] = *[p+6] = p[6]

%s of (c+10) is 17, strlen in int will be 2.

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assume starting address as 100

100+(84-73-1)=100+10=110

after which two characters are there resulting in the length of 2.

Answer 1

1st concept is -Strlen does not count null pointer .

Now strlen(17) is 2.