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Why this C program is crashing?

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# 1  
Old 04-26-2011
Why this C program is crashing?
Hi,

Why I am getting 'SIGSEGV' in the following code?
Code:
char* p="abcde";
printf("%s", 3[p]);     // Segmentation Fault (core dump)

Kindly help me to understand what exactly makes the program to crash or the reason for the crashing.
# 2  
Old 04-26-2011
Don't you mean p[3] instead of 3[p]? Also, p[3] will give you a single char, whereas you'll need a char * for the printf, so better use p+3 instead.
# 3  
Old 04-26-2011
Quote:
Originally Posted by royalibrahim
Kindly help me to understand what exactly makes the program to crash or the reason for the crashing.
Kindly help me understand what you were even trying to do here, I'm not sure. It's crashing because you're giving %s something that's not a string.
# 4  
Old 04-26-2011
Code:
char* p="abcde";
printf("%c", 3[p]);

should work fine. I believe that this is just a test of the fact that p[3] and 3[p] are supposed to behave the same way. He probably going to enter the obfuscated C contest or something like that. Smilie
# 5  
Old 04-26-2011
When you ask UNIX to reference memory you do not own, the OS generates a signal, SIGSEGV. This triggers a crash dump.

%s in printf assumes that what you give it is an address in memory that points to a nul-terminated array of characters. IT blindly goes ahead and tries to access a valid string. Instead it tried to access some unknown address outside of process memory, so it crashed.
This User Gave Thanks to jim mcnamara For This Post:
royalibrahim
# 6  
Old 04-26-2011
Quote:
Originally Posted by Perderabo
Code:
char* p="abcde";
printf("%c", 3[p]);

should work fine. I believe that this is just a test of the fact that p[3] and 3[p] are supposed to behave the same way. He probably going to enter the obfuscated C contest or something like that. Smilie
Hehehe. I was thinking the same thing.

A bit more info for those interested in this "feature": Question 6.11

And ... it gets worse:

Code:
$ cat acomm.c
#include <stdio.h>

int
main(int argc, char **argv) {
	char *s = "char *s";
	char *a[1][2][3];
	a[0][1][2] = s;

	(void) printf("%s\n", a[0][1][2]);
	(void) printf("%s\n", 2[1[0[a]]]);
	(void) printf("%s\n", 2[0[a][1]]);
	(void) printf("%s\n", 1[a[0]][2]);
	/* etc */

	return 0;
}


$ cc -Wall -pedantic acomm.c 
$ ./a.out 
char *s
char *s
char *s
char *s
$ cc --version
i686-apple-darwin8-gcc-4.0.1 (GCC) 4.0.1 (Apple Computer, Inc. build 5370)
Copyright (C) 2005 Free Software Foundation, Inc.
This is free software; see the source for copying conditions.  There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.

Regards,
Alister
This User Gave Thanks to alister For This Post:
vgersh99
# 7  
Old 04-27-2011
Quote:
Originally Posted by jim mcnamara
When you ask UNIX to reference memory you do not own, the OS generates a signal, SIGSEGV. This triggers a crash dump.

%s in printf assumes that what you give it is an address in memory that points to a nul-terminated array of characters. IT blindly goes ahead and tries to access a valid string. Instead it tried to access some unknown address outside of process memory, so it crashed.
Awesome !! Thanks a ton Smilie
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