Accessing null-terminated character

Question

Inspired by this question.

Code:

#include <stdio.h>

int main()
{
   char arr[] = "Hello";
   char *ptr = arr + 5;
   printf("%s\n",ptr);
}

In the above code, I have accessed null-terminated character.

So, What actually happens when accessing null terminated character in literal string? Is it Undefined behaviour?

By "accessing" you mean reading? Why would the null character be any different to any other character in the string? If you tried to write, went *beyond* the bounds of the string, then the effect is undefined. — cdarke, Jun 27 '18 at 05:50

Sourav Ghosh · Accepted Answer · 2018-06-27T05:54:09.333

Essentially, you're passing an empty string as the argument, so it should be treated as such.

For %s conversion specifier, with printf() family

[...]Characters from the array are written up to (but not including) the terminating null character.[...]

In your case, the null terminator happens to appear at the first element in the array, that's it.

Just for clarification, accessing a null-terminator is OK, accessing a NULL pointer is not OK, and they both are different things!!

score 4 · Answer 2 · answered Jun 27 '18 at 05:50

You are basically still accessing a null-terminated string.
It is just zero characters long, i.e. it does not contain anything to print.

Your code is basically the same as

 printf("");

Compare this, not duplicate but similar question:

score 1 · Answer 3 · answered Jun 27 '18 at 05:49

1

Nothing particular. A pointer to the null character is interpreted as a zero-length string by functions that expect a string.

answered Jun 27 '18 at 05:49

Matteo Italia

3 Answers3