Program hangs when encounters switch

Hi folks,

I've come across a peculiar behavior of the program written below:

Code:

#include<stdio.h>
#include<stdlib.h>

void removevowels(char* sen);

int main()
{
	char *sen = "Hello, Good Morning!!";
	printf("String before function call: %s\n", sen);
	removevowels(sen);
	printf("String after function call: %s\n", sen);

	return 0;
}

void removevowels(char* sen)
{
	char *p, *q;
	p = q = sen;
	while(*p != '\0'){
		printf("sen = %x\n", sen);
		printf("p = %x\n", p);
		printf("q = %x\n", q);
		printf("%s\n\n", p);

		switch(*p){
			case 'a': case 'e': case 'i': case 'o': case 'u':
			case 'A': case 'E': case 'I': case 'O': case 'U':
				p++;
				break;

			default:
				*q = *p;
				p++;
				q++;
				break;
		}

	}
	*q = '\0';
}

Upon compiling this program using MinGW (on Win XP), the compilation goes fine without errors but the program hangs when it is run.

The same is not happening if I compile and run using Borland C Compiler (on Win XP).

I added a few printf statements and found that it hangs at the switch statement.

Please share your thoughts on this.

Regards
maverix

If by "hangs" you means "throws a segmentation fault", then yes. (At least that's what I got compiling with MinGW on XP.) sen points to a string literal and cannot be modified. And so when you set q to sen, you can't modify q either, since q now points to a string literal. Declare sen as a char array, then strcpy your string literal in there.

Yes, I get a segmentation fault as well, but other than that, I'm not sure what the function is supposed to be doing?

Ah!! yes.
I didn't think on those lines.

I changed the declaration as below

Code:

char sen[] = "Hello, Good Morning!!";

and it ran fine... :-)

Thanks for your inputs...

regards
maverix

Originally Posted by Elysia

Yes, I get a segmentation fault as well, but other than that, I'm not sure what the function is supposed to be doing?

I was just writing a function to remove vowels from a string!!!

However, as I had mentioned earlier, the program in the first post works when compiled with Borland C compiler. Any specific reason for it to work. Was wondering why it (BCC) is letting to change a string literal.

regards
maverix

Some compilers store the string in a read-only section and some do not. There's no (AFAIK) standard on which to define this behavior.

Well whether a string literal is truly "read-only" is a matter for the implementation. For a start, it would require OS support, so if you tried it using a DOS compiler (even if running on XP), then all code and data is writeable, even where it's declared const.

It's just another example of you need to understand what the standard is telling you, rather than trying to figure out the rules from observations on a specific compiler. The standard in this instance just states that "strings" can be placed in read-only memory.

Another point of view is to never equate "works" with "bug-free". For any given implementation, there are plenty of places where you get lucky and it does what you want, but try the same thing on another implementation, and it will hand you your ass on a plate.