January 15, 2003
The documentation says that strings are not zero-terminated, and so you need to call append(0) to zero-terminate them before passing them to C functions.  But then it goes on to give this example:

str.append(0);
printf("the string is '%s'\n", (char *)str);

So what's up with passing the format string directly to printf, without appending a zero?


January 15, 2003
Kimberley Burchett  wrote:
> The documentation says that strings are not zero-terminated, and so you need to
> call append(0) to zero-terminate them before passing them to C functions.  But
> then it goes on to give this example:
> 
> str.append(0);
> printf("the string is '%s'\n", (char *)str);
> 
> So what's up with passing the format string directly to printf, without
> appending a zero?
> 
> 

When you give a constant string in D (like the format string you mentioned), D allocates a null terminator in the memory, even though the array doesn't stretch to include that character.  That is, when low-level C routines look at the memory, they will see the null terminator, even though it doesn't show up in the .length of any D array.

This is not the case, of course, for any dynamically generated string. You have to append to those.

A very perceptive question.  I hope that my answer makes sense.

January 15, 2003
"Russell Lewis" <spamhole-2001-07-16@deming-os.org> wrote in message news:3E258C2A.9040902@deming-os.org...
> A very perceptive question.  I hope that my answer makes sense.

Yes, it is a very good question, and your answer was spot on.


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