On 13/11/05, Paul <[EMAIL PROTECTED]> wrote:
> Hi,
>
> It's been a while since I've put on my programmer's hat, but I seem
> to remember some weirdness with streams and functions like this. I
> know it's a kludge, but if you create a var and assign read()'s value
> to the var and then output the var to cout does it work?

Nope, same error message. If try std::cout << read(int, NULL, int); it
compiles fine, but always returns a value of -1.

Looking at unistd.h (mentioned in the error) there is a function read():

<code>
/* Read NBYTES into BUF from FD.  Return the
   number read, -1 for errors or 0 for EOF.

   This function is a cancellation point and therefore not marked with
   __THROW.  */
extern ssize_t read (int __fd, void *__buf, size_t __nbytes);
</code>

I don't know a lot about the "inner workings" but it seems to me that
this function is getting preference over pngwriters read() - hence the
invalid conversion & missing arguments. But I always thought that if a
function was defined twice, you get told about it.

Dan

--
http://www.danicity.co.uk


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