Thread: Null pointer handling Linux vs Windows

Hi

I have a function that checks the value pointed by a string literal. In case of NULL pointer, function supposed to return NULL as return value.

When I compiled code in linux, I could not catch the NULL pointer with "static_str == 0" test condition but I had to write "static_str[0] == 0" which seems weird.

In Windows (Win 7, Mingw), the "static_str[0] == 0" did not work but "static_str == 0".

Is there a difference in handling of NULL pointer between OS's? Actually, the code I use in Win does more sense to me.

Below is a code snippet from the function

Code:

char *foo(char *str, const char *delim) {
	static char *static_str=0;		/* var to store last address */

	/* delimiter cannot be NULL
	* if no more char left, return NULL as well
	*/
	if (delim==0 || (str == 0 && static_str == 0))
		return 0;

Actually, I just copied the header file to my Win disk and compile the code again. No compiler error in Windows but program crashed. I opened the program in GDB and examined the function and realized that the program crashes at that point. Although, static_str[0] also does not makes sense to me, I cannot evaluate the cases where both str and static_str is set to NULL.

Please note, this is a strtok() like function that, after first call, static_str has the value retained from last call of the function foo().

Code:

#include <stdio.h>

char *zStrtok(char *str, const char *delim) {
	static char *static_str=0;		/* var to store last address */
	int index=0, strlength=0;		/* integers for indexes	*/
	int found = 0;				/* check if delim is found */

	/* delimiter cannot be NULL
	* if no more char left, return NULL as well
	*/
	if (delim==0 || (str == 0 && static_str == 0))
		return 0;

	if (str == 0)
		str = static_str;

        /* get length of string */
        while(str[strlength])
                strlength++;

	/* find the first occurance of delim */
	for (index=0;index<strlength;index++)
		if (str[index]==delim[0]) {
			found=1;
			break;
		}

	/* if delim is not contained in str, return str */
	if (!found) {
		static_str = 0;
		return str;
	}

	/* check for consecutive delimiters
	*if first char is delim, return delim
	*/
	if (str[0]==delim[0]) {
		static_str = (str + 1);
		return (char *)delim;
	}

	/* terminate the string
	* this assignmetn requires char[], so str has to
	* be char[] rather than *char which will cause SIGSEGV
	*/
	str[index] = '\0';

	/* save the rest of string */
	if ((str + index + 1)!=0)
		static_str = (str + index + 1);
	else
		static_str = 0;

		return str;
}

int main() {
char s[]="A,B,C,D,E,F,G,H,I";
printf("1  %s\n",zStrtok(s,","));
printf("2  %s\n",zStrtok(NULL,","));
printf("3  %s\n",zStrtok(NULL,","));
printf("4  %s\n",zStrtok(NULL,","));
printf("5  %s\n",zStrtok(NULL,","));
printf("6  %s\n",zStrtok(NULL,","));
printf("7  %s\n",zStrtok(NULL,","));
printf("8  %s\n",zStrtok(NULL,","));
printf("9  %s\n",zStrtok(NULL,","));
printf("10  %s\n",zStrtok(NULL,","));
printf("11  %s\n",zStrtok(NULL,","));
printf("12  %s\n",zStrtok(NULL,","));
printf("13  %s\n",zStrtok(NULL,","));
printf("14  %s\n",zStrtok(NULL,","));
return 0;
}

after 10th printf(), it is (null), if static_str[0] is used, program crashes in Win whereas, in linux, I had to add it to evaluate the case.

technically, they are the same, aren't they? I used the first convention for ease of code review.

Code:

str[index] = '\0'; != *(str+index) = '\0';

But what is the effect of this to the line 11 where the code crashes

Well, I will definitely try to snippet you gave. But before that, won't it give SIGSEGV before evaluating the condition in the code static_str != NULL && static_str[0] == '\0'?

Hmm....Ok, coming back to first question. Why it run in Linux fine but crashed under Win? There is simply a bug in the code, but why Linux did not crash ( well, this can't be because it is more robust OS than redmond OS :} )

Actually, I meant static_str[0] version (in line 11 at post #5)