Well... No.
If you compare a signed integer with a unsigned char, the char gets
promoted to a _signed_ integer, and the comparison is signed. It is NOT
a unsigned comparison.
And THIS is one example of why it gets complicated.
The C logic for type expansion is just a tad too easy to get wrong, and
the strict type-checking you normally have with well-written ANSI C simply
does not exist for integer types. The compiler will silently just do the
promotion..
Somebody mentioned -Wsign-compare. Try it with the example above. It won't
warn at all, exactly because under C both sides of such a compare have the
_same_ sign, even if one is a "unsigned char", and the other is a "signed
int".
Try it yourself if you don't believe me.
Please guys. The issue of sign in comparisons are a LOT more complicated
than most of you seem to think.
Linus
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