Convert String to Lower Case Crashed

[woozy]

Here I am again, and here is another question.

Again I'm trying to write replacement functions of exist ones in order to understand how C works. Here is my function, and it supposed to do a string conversion to lower cases:

int my_strlwr(char *str)
{
	int i;

	for (i = 0; *(str+i) != NULL; i++)
	{
		if (*(str+i) > 64 && *(str+i) < 91)
		{
			*(str+i) += 32;
		}
	}

	return 0;
}

The test code is:

int main()
{
	char *str = "Some";

	my_strlwr(str);

	printf("%s\n", str);

	return 0;
}

Somehow when the program executes to that if statement, it crashes. Please help, thanks in advance.

ps: and I know the program is written in a crappy manner, I'll make it neater after the major problem is solved.

[Dave_Sinkula]

This is not a writable string.

char *str = "Some";

This is.

char str[] = "Some";

[edit]Here

*(str+i) != NULL

oughtta be

*(str+i) != '\0'

[woozy]

Oh~~~damn it...Thank you so much Dave.

And btw, I thought NULL can represent '\0'

[Dave_Sinkula]

It can. It may be a pointer.

[woozy]

So '\0' in this case is more strict, right?

[Doodle77]

So '\0' in this case is more strict, right?

They are the same, but NULL is generally used as a value for pointers (it = 0 though)

Pointers can use array subscript notation, so

*(str+i) += 32;

is the same as

str[i] += 32;