Is `x-- > 0 && array[x]` well-defined behavior in C++?
Can I use x
on both sides of a boolean expression when I post-increment it on the left side?
The line in question is:
if(x-- > 0 && 开发者_JAVA技巧array[x]) { /* … use x … */ }
Is that defined through the standard? Will array[x] use the new value of x
or the old one?
It depends.
If &&
is the usual short-circuiting logical operator, then it's fine because there's a sequence point. array[x]
will use the new value.
If &&
is a user (or library) defined overloaded operator, then there is no short-circuit, and also no guarantee of a sequence point between the evaluation of x--
and the evaluation of array[x]
. This looks unlikely given your code, but without context it is not possible to say for sure. I think it's possible, with careful definition of array
, to arrange it that way.
This is why it's almost always a bad idea to overload operator&&
.
By the way, if ((x > 0) && array[--x])
has a very similar effect (again, assuming no operator overloading shenanigans), and in my opinion is clearer. The difference is whether or not x
gets decremented past 0, which you may or may not be relying on.
Yes, it is well defined. &&
introduces a sequence point.
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