CODEWITHSUNDEEP
c
Vignesh Radhan
Vignesh Radhan

Reputation: 11

What will be the output of printf("%s",*&*str); where in str is a pointer?

this was the code snippet asked for me in a interview and pls explain me what is the use of &.

#include<stdio.h>
int main()
{
char *str="INCLUDEHELP";
printf("%s",*&*str);
}

Upvotes: 0

Views: 276

Answers (2)

KamilCuk
KamilCuk

Reputation: 141085

The &* is an empty operation. The &*str (or like &*&*&*&*&*&*&*&*&*str or any number of &*) is equivalent to just str. See C11 note102. We can omit it.

char *str="INCLUDEHELP";
printf("%s", *str);

This code with %s is invalid. The %s expects zero terminated char array. The *str is a char with the value of 'I', a character. The program on linux-like systems will most probably receive a segmentation fault signal, because 'I' will be an invalid address of a zero terminated character array.

If it were:

char *str="INCLUDEHELP";
printf("%c", *str);

or:

#include<stdio.h>
int main()
{
char *str="INCLUDEHELP";
printf("%c",*&*str);
}

, then the program would print a single character I on it's stdout.

Upvotes: 2

0___________
0___________

Reputation: 67546

The output is most probably the segfault.as it is the Undefined Behaviour. The &*... from the right mean. Dereference the pointer, get the address of the dereferenced char, and dereference this address again passing the char instead of the pointer to printf.

Upvotes: 0

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