Strings in C Programming Language

Question

#include <stdio.h>

int main() {
    char str3[6] = {'h', 'e', 'l', 'l', 'o', '\0'};
    char str4[ ] = {'h', 'e', 'l', 'l', 'o'}; 
    printf("%s, %s", str3, str4);
    
    return 0;
}

The output for this is : hello, hellohello

Why two times hello ????

As an experiment, try defining `str4` before `str3`. You will probably get the output `hello, hello` or a segmentation violation. — Ian Abbott, Feb 26 '21 at 12:31
@IanAbbott I tried ```str4``` before ```str3``` but it just gives the same output. — Deva Narayana Babu, Feb 27 '21 at 04:18

score 1 · Accepted Answer · answered Feb 26 '21 at 12:28

1

str4 has no 0-terminator, so printf keeps reading out of bounds of the array and by luck (or not) there seems to be str3 directly behind str4, so it reads till the 0-terminator of str3.

In the end ==> undefined behavior, don't rely on that.

answered Feb 26 '21 at 12:28

mch

9,424
2
28
42

ooh, i see, but even when i change the order of their output i.e. ```str4``` before ```str3``` the same output is being displayed. – Deva Narayana Babu Feb 27 '21 at 04:20
Yes, there is no guarantee for the order of the variables in the memory. – mch Feb 27 '21 at 08:43

score 0 · Answer 2 · answered Feb 26 '21 at 12:29

0

str4 does not contain a null terminator. Since str4 and str3 are sequentially allocated on the stack by the compiler, the second %s in the printf will start printing str4 and due to the lack of a null terminator continue with printing str3.

answered Feb 26 '21 at 12:29

Papachristodoulopoulos

1
1

Strings in C Programming Language

2 Answers2