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by kragen
330 days ago
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The usual meaning of "data path" https://en.wikipedia.org/wiki/Datapath is the path from the register file (and/or memory access unit) to the ALU and back to the register file (and/or memory access unit). So we could say that both the 8088 and the 68000 had a 16-bit data path, because they used 16-bit buses for those transfers and a 16-bit ALU, even though the 68000 had 32-bit registers and the 8088 had an 8-bit data bus to connect it to RAM. The 68020 implemented the same instructions and registers as the 68000 (and additional ones) but used a 32-bit data path, so they were twice as fast. In what sense does a virtual machine instruction set architecture with no hardware implementation have a "data path" separate from its arithmetic size? You seem to be using the term in a nonstandard way, which is fine, but I cannot guess what it is. By your other criteria, the (uncontroversially "16-bit") 8088 would be an 8-bit computer, except that it had a 20-bit address space. |
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