3 ms·
It's not a matter of address lines; registers are 16-bit (can addresss only 64 kilobytes), so you need segmentation regardless of whether your segements are map
by a1369209993 6y ago
It's not a matter of address lines; registers are 16-bit (can addresss only 64 kilobytes), so you need segmentation regardless of whether your segements are mapped into 1MB, 16MB, or 4GB. (For Abus = Areg + Sreg << 4, 8, or 16 respectively.)
- ibiza 6y agoMotorola jumped from their 8-bit 6809 to a full 32-bit (internal) processor in 1979. Intel delivered a 32-bit processor 6 years later in 1985. Intel Motorola 1974 8080 8-bit 1974 6800 8-bit 1976 8085 8-bit 1978 6809 8-bit 1978 8086 16-bit 1979 68000 32-bit 1982 80286 16-bit 1985 80386 32-bit
- a1369209993 6y agoYes, but that's not a matter of "just add[ing] some address lines"; it's a matter of adding additional bits to the registers (and probably the ALU and internal data buses, but the registers are the limiting factor: if you don't have 32-bit registers you can't have (32-bit) flat address space). Even if it were "just", it would be "just add some register bits".