5/25/2023 0 Comments Dclock 8088To figure out how long a sequence of instructions will take, one would need to figure out when these operations will occur, and then add to every instruction's execution time the amount of time it would spend waiting either for an instruction to be fetched, or for the bus unit to finish a prefetch operation. Īny time the main CPU isn't asking for a memory access and there fewer than four bytes (8088) or three two-byte words (8086) of code have been fetched ahead of the current execution point, the prefetch unit will initiate a code fetch, which will take four cycles. accessing display-card memory, or when a DRAM refresh occurs. Every memory cycle on the 8088 or 8086 takes a minimum of four cycles, and on most machines they will usually take exactly four cycles except when e.g. Generally, on the 8088 execution speed will be limited by memory access, while the 8086 will be better balanced. ![]() On the 80, execution involves two parallel processes-memory access and internal computation-and will be limited by the speed of whichever is slower. Even if your assembler copes with it, it may introduce incompatibilities and hard to correct problems when using a different environment. *2 - Please do not use white-spaces within parameters. *1 - If your question is about an 8088, as used in the original PC, then it's actually 29, as described here. How I can calculate the duration of this program?Īs usual by adding up the clock cycles and multiply them by the clock period used. What is a 'Wait tick'? Mind to specify it, preferably in CPU clocks? Is that memory cycle or CPU clock? These are not the same and bus clock is an ambiguous term here.Īccess to memory is done without WAIT, and to I/O with one WAIT tick. In real world application the queue is rarely empty, as only a few high performance instructions can empty it when executed in sequence.īase line: Prefetch wait is usually zero and thus doesn't matter, and if it does, one needs to analyze the instructions before that sequence. Since move's 13 clocks leaves a lot of room (only 4 are needed to get the data), the OUT usually can be fetched in parallel, leaving the total penalty here at 0 or 4 CPU clocks. ![]() Depending on the prior instructions they could be already loaded, adding zero cycles, or 4 additional cycles per fetch, adding 4 or 8 cycles total. ![]() Above code is 3 bytes (8A 07 EE), so two code fetches are necessary. In addition it depends on the prefetch queue being filled or not, as code fetches are additional. Since it's a byte access, no penalty for misalignment can happen. To start with basic timing for 8086 (*1) in CPU clocks is 21 clock cycles.ġ3 cycles, consisting of 8 cycles for the instruction (MOV reg8,mem) plus 5 to calculate the address from. I have the following assembly code for 8086
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