On Thursday 28 July 2011 06:22:10 Gene Buckle wrote:
> On Thu, 28 Jul 2011, Jari Häkkinen wrote:
> >>> Are you sure about that? I just tried it with a little example and
> >>> at least gcc compiles both variants to the exact same assembly
> >>> code. Tried it with and without -O2.
> >>
> >> That would freak me out. Doesn't "++j" mean "increment j, then test"
> >> whereas "j++" means "test j, then increment"?
> >
> > No, for a for loop
> >
> > for ( [1]; [2]; [3] )
> >
> > where [3] is ++j will increment j before use. However, in an
> > if-statement the complete statement [3] is evaluated before the test [2]
> > is done. If the compiler is smart it will produce the fastest binary
> > code regardless ++j or j++. However, if the [3] is more complicated like
> > a hypothetical i = ++j + k the compiler will most probably generate
> > different binary code (compared to i = ++j + k).
>
> Right, but j++ will increment _after_ it's used, correct? So how could
> ++j vs j++ generate the same assembly code and be correct?
Because the for loop is a case where it simply doesn't matter. Translate the
for to a while:
int i = 0;
while (i < 10)
i++;
is just the same as:
int i = 0;
while (i < 10)
++i;
because the value of the incrementing expression is not used anywhere.
The compiler recognizes this as one of the simplest cases of optimization.
Stefan
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