Re:So you think you know C?
by ekiru on Tuesday November 08, 2016 @08:25PM
(#12905408)
char is also allowed to be a signed type, in which case overflow is undefined behavior.
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Re:So you think you know C?
by evincarofautumn on Tuesday November 08, 2016 @10:50PM
(#12905914)
It’s not because of the signedness of char—not only do character literals have type int, but even if this weren’t a literal, the char would be promoted to (signed) int before the multiplication. But of course the multiplication of two ints can still overflow.