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Re: Would this cause undefined behavior please?

* * *z=++x||++y && ++z;
means the same thing as:
* * *z = ++x || (++y && ++z);

To evaluate the result of the || operator,
(++x) is evaluated first, like you said,
but after that, there is nothing more to do.

(++y && ++z) is not evaluated.
Of course! Many thanks.
Aug 18 '08 #1
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