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int x,y,z;

x=y=z=1;

z=++x||++y && ++z;

My understanding is that the order of things here are:

++x = x = 2

As x is 'true', the ++y doesn't operate

To evaluate &&, ++z operates, setting z to 2.

true && true, and that assigns 1 to z.

Is this right 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.

pete <> wrote:
>
> (++y && ++z) is not evaluated.

Which is good because, if it were, you (the OP, not pete) would be
trying to set the value of z twice in the same expression without a
sequence point between them, which is undefined behavior, just like:

z = ++z;
--
Larry Jones

I just can't identify with that kind of work ethic. -- Calvin