Made Easy Test Series:Programming-String

Question

#include<stdio.h>
#include<string.h>
int main(void) {
char *a="MADEEASY";
char *b="GATECSIT2019";
char *r=a;
char *s=b;
printf("%d",(int)strlen(b+3[r]-1[s]));
return 0;
}
 

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Ans is given $8.$ I know basic thing, but couldnot getting what strlen(b+3[r]-1[s]) returning?? Plz explain.

Comments

@amitqy

 

What is the answer?

Im getting 6

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Is it 8?

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yes.how printf statement is working

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its 8

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Here we need to see the execution steps to understand how it is 8 :-

strlen(b + 3[r] - 1[s])

what is 3[r] ? it is *(r+3) which is  E.

and 1[s] is *(1+s) which is A.

Here subtraction of characters give 4 since E-A is 4 (positional difference in ASCII).

now b + 4 will be considered from position 4 .

strlen('CSIT2019') is 8

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char *a =  "GATE";

printf("%s",a)--->outputs GATE

printf("%d",a)--->outputs starting starting address of string "GATE"

correct?

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8 is answer

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Correct

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8 is the correct answer

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Same question: https://gateoverflow.in/311306/made-easy-test-series-programming-string

Answer 1

strlen(b+3[r]-1[s]) :

first: 3[r] is same as r[3] i.e., 'E'

and 1[s] is same as s[1] i.e., 'A'

3[r] - 1[s] = 'E' - 'A' = 4

now, b + 4 refers to  4 th index in b i.e, 'C' (starting position)

strlen(b+4) = strlen("CSIT2019") = 8

Thus, it will return 8.

correct me if i am wrong.

Comments

But Arithmetic operator is Left to right associativity.

so why you evaluate the right first ?

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yes, but it won't differ.

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b+3[r]-1[s]=b+*(r+3)-*(s+1)

                   =b+E-A

                    =b+ 69 -65

                     =b+4

                       =  strlen(b+4)

                         =strlen(CSIT2019)

                          =8   { ASCII of A=65 &E=69}

Answer 2

b+3[r]-1[s]=b+*(r+3)-*(s+1)

                   =b+E-A      {ASCII OF A=65 &E=69}

                    =b+ 69 -65

                     =b+4

                       =  strlen(b+4)

                         =strlen(CSIT2019)

                          =8               (So answer is 8)

Answer 3

b+3[r]-1[s]=b+*(r+3)-*(s+1)    (r[3]=*(r+3)=*(3+r)=3[r])

                   =b+E-A      {ASCII OF A=65 &E=69}

                    =b+ 69 -65

                     =b+4 (here b points to C right now)

                       =  strlen(b+4)

                         =strlen(CSIT2019) (strlen never count \0')

                          =8               (So answer is 8)