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Why doesn't the compiler emit a warning for all UB it finds while compiling? Or do regular programs rely on this too much to be feasible? It must be something l
by MichaelGG 11y ago
Why doesn't the compiler emit a warning for all UB it finds while compiling? Or do regular programs rely on this too much to be feasible? It must be something like that, or there'd not be much performance gains to be had by exploiting UB right?
- DanWaterworth 11y agoIt's not necessarily detectable statically.
- lmm 11y agoAny time you add two (signed) integers, that's potentially undefined behaviour.