Not Null Terminated character array

Question

#include <stdio.h>

char s[3] = "Robert";
int main()
{
    printf("%s",s);
}

Output: Rob

How does this get printed properly? The string is not null terminated. I saw the assembly. It used .ascii for storing "Rob" which is not null terminated. I expected some garbage along with Rob to be printed. Can someone explain me this behaviour?

Probably because you got lucky and the next byte in memory after s is \x00. — Paulo Bu
– Paulo Bu, Commented Apr 26, 2014 at 22:18
I am just trying to learn how character arrays and literals are stored. — Anup Buchke
– Anup Buchke, Commented Apr 26, 2014 at 22:19
FWIW, this was allowed in C. C++11 § C.1.7 [diff.decl] says In C++, when initializing an array of character with a string, the number of characters in the string (including the terminating ’\0’) must not exceed the number of elements in the array. In C, an array can be initialized with a string even if the array is not large enough to contain the string-terminating ’\0’ — Qaz
– Qaz, Commented Apr 26, 2014 at 22:25
Thanks Chris for pointing to the rules.. I tried out in c++. It does give error. g++ enforces it. The more I learn about about c and c++ , i realize that they are different in many aspects. — Anup Buchke
– Anup Buchke, Commented Apr 26, 2014 at 22:29

rkhb · Accepted Answer · 2014-04-27 09:57:24Z

4

Your "Rob" has been stored in an extra section of the executable. The sections in an executable are aligned, i.e. the section with the data is padded with 0 until the next section. So printf got "its" 0 from the section padding. To illustrate:

#include <stdio.h>

char dummy[] = "ocop";
char s[3] = "Robert";
char second[] = "in Hood";
int main( void )
{
    printf("%s",s);
    return 0;
}

Output (MinGW-GCC w/o optimization): Robin Hood
Output (MinGW-GCC with optimization): Robocop

Now there is no 0 from the padding but the begin of the next string which will be outputted as well.

answered Apr 27, 2014 at 9:57

rkhb

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