Or it will run until you run out of ram or rom space, assuming you can access all of the ram/rom space you are pointing to.<div>...</div><div>Just thought of something, the code does not say while((*x++ <b><font class="Apple-style-span" color="#ff0000">==</font></b> *y++)){},</div>
<div>thus the value y is pointing to, incremented by one, should always be able to <b>successfully </b>be stored in the register x is pointing to, even if the value in *x is incremented before the store.</div><div>Thus == to while(1){}</div>
<div><br></div><div>Cheers,</div><div>Pieter.<br><br><div class="gmail_quote">On Thu, Aug 4, 2011 at 8:53 AM, Andre Renaud <span dir="ltr"><<a href="mailto:andre@bluewatersys.com">andre@bluewatersys.com</a>></span> wrote:<br>
<blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex;"><div class="im">On 03/08/11 22:13, Volker Kuhlmann wrote:<br>
> On Wed 03 Aug 2011 11:40:03 NZST +1200, Charles Manning wrote:<br>
><br>
>> In my books there is also a very hot place reserved for those who write things<br>
>> like<br>
>> while((*x++ = *y++)){}<br>
><br>
> You can't really compare this with previous examples. This one is much<br>
> different because it is undefined, unpredictable and therefore useless<br>
> and much worse. Your choc fish challenge was defined in its behaviour<br>
> and predictable, although extremely bad programming and a convoluted way<br>
> to say while(1); .<br>
<br>
</div>I don't think this is undefined - isn't this a simple implementation of<br>
strcpy (assuming x & y are char *)? It will continue around until it<br>
assigns a 0 pointer, at which point it will stop. It would be undefined<br>
if you had *x++ = *x++.<br>
<br>
Regards,<br>
<font color="#888888">Andre<br>
</font><div><div></div><div class="h5"><br>
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