• Buffalox@lemmy.world
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      2 months ago

      It means pointer width.

      https://en.wikipedia.org/wiki/64-bit_computing

      64-bit integers, memory addresses, or other data units[a] are those that are 64 bits wide. Also, 64-bit central processing units (CPU) and arithmetic logic units (ALU) are those that are based on processor registers, address buses, or data buses of that size.

      It also states Address bus, but as I mentioned before, that doesn’t exist. So it boils down to instruction set as a whole requiring 64 bit processor registers and Databus.
      Obviously 64 bits means registers are 64 bit, the addresses are therefore also 64 bit, otherwise it would require type casting every time you need to make calculations on them. But it’s the ability to handle 64 bit registers in general that counts, not the address registers. which is merely a byproduct.

    • Buffalox@lemmy.world
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      2 months ago

      It means pointer width.

      Where did you get that from? Because that’s false, please show me dokumentation for that.
      64 bit always meant the ability to handle 64 bit wide instructions, and because the architecture is 64 bit, the pointers INTERNALLY are 64 bit, but effectively they are only for instance 40 bit when accessing data.
      Your claim about pointer width simply doesn’t make any sense.
      That the CPU should be called by a single aspect they can’t actually handle!!! That’s moronic.