Programming Forums > Application Development > C

SnakeByte · Mar 11th, 2008, 12:17 PM

Hi, I'm trying to learn C coding and I happen to be at a chapter in a book I'm reading which deals with segmentation faults. What I'm trying to do is get a program, that accepts command-line arguments, to crash (with a seg fault) when it isn't given any arguments. The code from the book is as follows:

convert2.c

 C Syntax (Toggle Plain Text) 
#include <stdio.h>
 
void usage(char *program_name) { 
	printf("Usage: %s <message> <# of times to repeat\n", program_name);
	exit(1);
}
 
int main(int argc, char *argv[]) {
	int i, count;
 
	//if(argc < 3)		// If fewer than 3 arguements are used,
	//	usage(argv[0]); // Display usage messages and exit.
 
	count = atoi(argv[2]); //Convert the 2nd arg into an integer.
	printf("Repeating %d times...\n", count);
 
	for(i=0; i < count; i++)
		printf("%3d - %s\n", i, argv[1]); // Print the 1st arg.
}
#include <stdio.h>

void usage(char *program_name) { 
	printf("Usage: %s <message> <# of times to repeat\n", program_name);
	exit(1);
}

int main(int argc, char *argv[]) {
	int i, count;

	//if(argc < 3)		// If fewer than 3 arguements are used,
	//	usage(argv[0]); // Display usage messages and exit.
	
	count = atoi(argv[2]); //Convert the 2nd arg into an integer.
	printf("Repeating %d times...\n", count);

	for(i=0; i < count; i++)
		printf("%3d - %s\n", i, argv[1]); // Print the 1st arg.
}

When I compile this code in gcc (v4.1.3) on Ubuntu 7.10 I get this message:
$ gcc -g -o convert2 convert2.c
convert2.c: In function ‘usage’:
convert2.c:5: warning: incompatible implicit declaration of built-in function ‘exit’

It still compiles, but when I run it, it does this:
$ ./convert2
Repeating 0 times...

Since I'm trying to break the program, this isn't the result I want. Does anyone know how to "break" this program so that it will cause a seg fault?

BTW, the desired result would be this:
$ ./convert2
Segmentation fault (core dumped)
$

Narue · Mar 11th, 2008, 1:06 PM

>What I'm trying to do is get a program, that accepts command-line
>arguments, to crash (with a seg fault) when it isn't given any arguments.
What you're seeing is undefined behavior. It might crash, it might not. In this case, accessing argv[2] doesn't touch memory outside of your address space, and when atoi fails, it returns 0. Both of these are undefined behavior, but you lucked out in that it doesn't cause the error you were expecting.

The only way to guarantee a seg fault is to raise it yourself.

mbd · Mar 11th, 2008, 2:21 PM

(Toggle Plain Text)

#include <stdlib.h>

will get rid of the warning about exit being implicitly defined

to get a seg fault you should just try writing to a specific memory address. something along the lines of:

(Toggle Plain Text)

int *i;
i = 0;
*i = 0;

i think that would generate a seg fault.

Ancient Dragon · Mar 12th, 2008, 12:00 AM

There are lots of things that will generate seg faults -- division by 0 if another one of them

(Toggle Plain Text)

int x = 123 / 0;

Jimbo · Mar 12th, 2008, 1:08 AM

Out of curiosity, any particular reason you want to cause a seg fault? It is an odd request...

L7Sqr · Mar 12th, 2008, 6:26 AM

(Toggle Plain Text)

#include <signal.h>

/* ... */

raise (SIGSEGV);

Jessehk · Mar 12th, 2008, 7:50 AM

Quote:

Originally Posted by L7Sqr

(Toggle Plain Text)

#include <signal.h>

/* ... */

raise (SIGSEGV);

That's cheating.

SnakeByte · Mar 12th, 2008, 7:55 AM

Quote:

Out of curiosity, any particular reason you want to cause a seg fault? It is an odd request...

Jimbo: I'm trying to cause a seg fault because the particular book I'm using to learn C attempts to explain the reason seg faults occur through debugging (gdb) and looking at memory addresses. In order to understand why/how they're caused, the book gives you a program that should cause a segmentation fault.

Ancient Dragon · Mar 12th, 2008, 9:06 AM

This sounds like a good idea for a game -- 100 ways to create a seg fault

mbd · Mar 12th, 2008, 11:24 AM

Quote:

Originally Posted by Ancient Dragon

There are lots of things that will generate seg faults -- division by 0 if another one of them

(Toggle Plain Text)

int x = 123 / 0;

a segmentation fault is only caused by accessing memory in a way which is not allowed. division by zero is a different type of exception.

Mar 11th, 2008, 12:17 PM	#1
SnakeByte Newbie Join Date: Mar 2008 Posts: 2 Rep Power: 0	Causing a Seg Fault on Purpose Hi, I'm trying to learn C coding and I happen to be at a chapter in a book I'm reading which deals with segmentation faults. What I'm trying to do is get a program, that accepts command-line arguments, to crash (with a seg fault) when it isn't given any arguments. The code from the book is as follows: convert2.c C Syntax (Toggle Plain Text) #include <stdio.h> void usage(char program_name) { printf("Usage: %s <message> <# of times to repeat\n", program_name); exit(1); } int main(int argc, char argv[]) { int i, count; //if(argc < 3) // If fewer than 3 arguements are used, // usage(argv[0]); // Display usage messages and exit. count = atoi(argv[2]); //Convert the 2nd arg into an integer. printf("Repeating %d times...\n", count); for(i=0; i < count; i++) printf("%3d - %s\n", i, argv[1]); // Print the 1st arg. } #include <stdio.h> void usage(char program_name) { printf("Usage: %s <message> <# of times to repeat\n", program_name); exit(1); } int main(int argc, char argv[]) { int i, count; //if(argc < 3) // If fewer than 3 arguements are used, // usage(argv[0]); // Display usage messages and exit. count = atoi(argv[2]); //Convert the 2nd arg into an integer. printf("Repeating %d times...\n", count); for(i=0; i < count; i++) printf("%3d - %s\n", i, argv[1]); // Print the 1st arg. } When I compile this code in gcc (v4.1.3) on Ubuntu 7.10 I get this message: $ gcc -g -o convert2 convert2.c convert2.c: In function ‘usage’: convert2.c:5: warning: incompatible implicit declaration of built-in function ‘exit’ It still compiles, but when I run it, it does this: $ ./convert2 Repeating 0 times... Since I'm trying to break the program, this isn't the result I want. Does anyone know how to "break" this program so that it will cause a seg fault? BTW, the desired result would be this: $ ./convert2 Segmentation fault (core dumped) $ __________________ Boy, it sure would be nice if we had some grenades, don't you think? - Jayne Cobb Last edited by SnakeByte; Mar 11th, 2008 at 12:33 PM.

Mar 11th, 2008, 1:06 PM	#2
Narue Professional Programmer Join Date: Sep 2005 Posts: 419 Rep Power: 3	Re: Causing a Seg Fault on Purpose >What I'm trying to do is get a program, that accepts command-line >arguments, to crash (with a seg fault) when it isn't given any arguments. What you're seeing is undefined behavior. It might crash, it might not. In this case, accessing argv[2] doesn't touch memory outside of your address space, and when atoi fails, it returns 0. Both of these are undefined behavior, but you lucked out in that it doesn't cause the error you were expecting. The only way to guarantee a seg fault is to raise it yourself. __________________ Even if the voices aren't real, they have some pretty good ideas.

Mar 11th, 2008, 2:21 PM	#3
mbd Programmer Join Date: Nov 2007 Posts: 86 Rep Power: 1	Re: Causing a Seg Fault on Purpose (Toggle Plain Text) #include <stdlib.h> #include <stdlib.h> will get rid of the warning about exit being implicitly defined to get a seg fault you should just try writing to a specific memory address. something along the lines of: (Toggle Plain Text) int i; i = 0; i = 0; int i; i = 0; i = 0; i think that would generate a seg fault.

Mar 12th, 2008, 12:00 AM	#4
Ancient Dragon PFO God In Training Join Date: Jun 2005 Location: near St Louis, MO. (USA) Posts: 532 Rep Power: 4	Re: Causing a Seg Fault on Purpose There are lots of things that will generate seg faults -- division by 0 if another one of them (Toggle Plain Text) int x = 123 / 0; int x = 123 / 0; __________________ I Like Ike. Vote for Dwight Eisenhower this November. --This message brought to you by the the Procrastinators Club Of America.

Mar 12th, 2008, 1:08 AM	#5
Jimbo Battle Programmer Join Date: Feb 2006 Location: Bellevue, WA, USA Posts: 754 Rep Power: 3	Re: Causing a Seg Fault on Purpose Out of curiosity, any particular reason you want to cause a seg fault? It is an odd request... __________________ <insert disclaimer here> <insert shameless plug for Visual Studio here>

Mar 12th, 2008, 6:26 AM	#6
L7Sqr Hobbyist Programmer Join Date: Jun 2005 Location: here Posts: 124 Rep Power: 0	Re: Causing a Seg Fault on Purpose (Toggle Plain Text) #include <signal.h> /* ... / raise (SIGSEGV); #include <signal.h> / ... */ raise (SIGSEGV); __________________ "...and though our kids are blessed their parents let them shoulder all the blame." - The Quiet Things That No One Ever Knows [BrandNew]

Mar 12th, 2008, 9:06 AM	#9
Ancient Dragon PFO God In Training Join Date: Jun 2005 Location: near St Louis, MO. (USA) Posts: 532 Rep Power: 4	Re: Causing a Seg Fault on Purpose This sounds like a good idea for a game -- 100 ways to create a seg fault __________________ I Like Ike. Vote for Dwight Eisenhower this November. --This message brought to you by the the Procrastinators Club Of America.

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Programming Forums > Application Development > C

Causing a Seg Fault on Purpose