Reply To: Execution

As far as the specializer is concerned, a call is just a zero-latency op. F(G(), x*y) has the same constraints as F(a+b, x*y). It can schedule the {call G, add} any time in relation to the {mil}, but both have to be done before {call F} can issue. Becuase it knows all the latencies, it tries to have everything ready to go and already dropped to the belt when it is needed. In the case of the nested call, recall that even with cascading both calls are in the same instruction, and multiply drops between readerPhase and opPhase, which is before callPhase. So for best speed, the mul must retire before either call, cascaded or not. Remember, both calls happen in one of the consecutive call phases of the single instruction; unlike other phases, there can be several call phases.

It doesn’t matter how long either function takes to execute; for scheduling, all calls take zero cycles, i.e. they complete in the same instruction that they issue in.