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main()

{

int i, n;

char *x = “girl”;

n = strlen(x);

*x = x[n];

for(i=0; i<n; ++i)

{

printf(“%s\n”,x);

x++;

}

}

Answers were Sorted based on User's Feedback



main() { int i, n; char *x = “girl”; n = strlen(x); ..

Answer / susie

Answer :

(blank space)

irl

rl

l

Explanation:

Here a string (a pointer to char) is initialized with a
value “girl”. The strlen function returns the length of the
string, thus n has a value 4. The next statement assigns
value at the nth location (‘\0’) to the first location. Now
the string becomes “\0irl” . Now the printf statement prints
the string after each iteration it increments it starting
position. Loop starts from 0 to 4. The first time x[0] =
‘\0’ hence it prints nothing and pointer value is
incremented. The second time it prints from x[1] i.e “irl”
and the third time it prints “rl” and the last time it
prints “l” and the loop terminates.

Is This Answer Correct ?    8 Yes 3 No

main() { int i, n; char *x = “girl”; n = strlen(x); ..

Answer / arhant jain

Segmentation fault

Is This Answer Correct ?    1 Yes 0 No

Post New Answer

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