On 14/01/2013 14:10, Jonas Maebe wrote:
That said, the compiler contains an optimization pass that tries to
remove 64 bit widenings on 32 bit platforms in case it turns out they
were not necessary. It will also catch the above example and the
generated code will only contain a 32 bit operations. Semantically,
the result of that expression remains 64 bit though.
So I am right about the 64 bits not needed.
And the hint is simply because the compiler (for keeping it simple) will
first extend *all* mixed signed/unsigned to 64 bit (even if not needed
in the end).
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