On 01/06/2026 00:54, Keith Thompson wrote:[...]
(INT_MIN + INT_MAX) + 1 is well defined. (INT_MIN + INT_MAX) +1
is equivalent, and is also well defined. INT_MIN + (INT_MAX +1)
has undefined behavior.
Oops, I forgot to delete some parentheses. I meant to write that
INT_MIN + INT_MAX + 1 is equivalent to (INT_MIN + INT_MAX) + 1.
The redundant parentheses don't impose any change in semantics.
Compilers can re-arrange integer arithmetic, despite new overflows, if
they know the result is the same. On pretty much any current
processor, a compiler generating code for integer "a + b + c" could do
the additions in any order - treating the operations as commutative
and fully associative. The final result will be the same in every
case where the original expression did not overflow (i.e., every case
with defined behaviour).
Right, good point. Since (INT_MIN + INT_MAX) + 1 is well defined,
if a compiler rearranges it so it evaluates INT_MAX + 1 as an
intermediate result, that's permitted **if** the result is the same.
(It makes more sense if the operands are variables with those values
rather than constants.) A compiler can take advantage of how the
hardware works. UB applies to the C source code, not (necessarily)
to the operations that are performed by the actual machine. If it
generates code that yields the correct result by consulting a Ouija
board, that's still conforming.
[...]
--
Keith Thompson (The_Other_Keith) Keith.S.Thompson+u@gmail.com
void Void(void) { Void(); } /* The recursive call of the void */
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