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It breaks binary compatibility. The compiler substitutes some value (e.g. the broken value), so if the sin(x) evaluation changes (to the correct one), then bina
by darkmighty 12y ago
It breaks binary compatibility. The compiler substitutes some value (e.g. the broken value), so if the sin(x) evaluation changes (to the correct one), then binary execution will fail.
- DannyBee 12y agoWhy would the compiler substitute the broken value unless the compiler was broken? Compilers generally don't use specialized local architecture instructions to compute their results, because they can't (It would break cross-compiling or targeting different platforms in a lot of cases) In fact, it's more likely the other way around: Compiler substitutes proper value. actual calls on sin(x) on the local platform give wrong answer.