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On 10/28/2024 9:56 PM, Richard Damon wrote:Of course. For that it needs its own code. This doesn't need to beOn 10/28/24 9:09 PM, olcott wrote:When HHH (unknowingly) emulates itself emulating DDD this emulated HHHOn 10/28/2024 6:56 PM, Richard Damon wrote:Then how did it convert the call HHH into an emulation of DDD again?>At machine address 0000217a HHH emulates itself emulating DDD without
It is IMPOSSIBLE to emulate DDD per the x86 semantics without the
code for HHH, so it needs to be part of the input.
knowing that it is emulating itself.
is going to freaking emulate DDD.
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