c gurus: whats the diff? "var[]" and "*var"?

xhi · 03-27-2007, 09:18 AM

> FYI "unspecified" is a code word for "It may or may not work

99% of the time, attempting to modify a literal will result in a segfault, you should declare literal declarations as const to help prevent modifying them.

Quote:

char * should point at some previously allocated data.
char [] will have space allocated in your heap or stack space.

i would word it, char[] cannot be null and char* could or just as well could not point to previously allocated data (as a * is null just as often as it is not)

kourama · 04-03-2007, 10:28 AM

Quote:

Originally Posted by jim mcnamara

This is not good. Nobody mentioned the C99 standard and what it says:

1. char var[]="abc";
This creates an array of char which is modifable
2. char *var="abc";
This creates an array of char, the results of modifying it are unspecified.

FYI "unspecified" is a code word for "It may or may not work. But don't expect to be able to port the code just because some compiler writer was nice enough to make it work for your platform."
In short "don't modify it"

3. char *var=NULL; [ or char *var; ] is a pointer to char and is modifable.

Although somewhat tangent, that's an excellent point. I have a copy of a recent C standards document somewhere. I oughta look at it every now and then!

Apologies for post<quote etiquette violation.