4 ms·
They say that it's a dynamic translation layer, so presumably it doesn't make a distinction about where the code came for (except maybe you'll lose some perform
by poizan42 9y ago
They say that it's a dynamic translation layer, so presumably it doesn't make a distinction about where the code came for (except maybe you'll lose some performance optimization if they are caching translated code)
- tmccrmck 9y agoMost dynamic emulators don't do instruction by instruction emulation - they take in a whole block of instructions and translate that block, caching chunks that may be used without any inference (eg. branching). Once you hit a similar instruction block you don't have to do any translation; just execute from the cache. This is why you have a load time for reading from an x86 image.