8位和16位处理器如何通过两个寄存器访问更多RAM? [英] How do 8-bit and 16-bit processors access more RAM with two registers?

问题描述

让我一直感到困惑的是8位计算机如何访问超过256字节的RAM.我知道它必须使用两个寄存器，但是任何一个都可以给我展示一下汇编代码中的示例吗?

Something that has always confused me is how 8-bit computers access more than 256 bytes of RAM. I know that it must use two registers, but can any one show me an example of what this would look like in assembly code?

赞:

mov a, [x]   ???

推荐答案

让我们假设我们在寄存器L和H中具有地址的LOWER和HIGHER 8bit一半.例如，我们要从地址32770读取字节dec = 8002 hex

Let's imagine we have LOWER and HIGHER 8bit half of the address in registers L and H. For example, we want to read byte from address 32770 dec = 8002 hex.

mov l, 02h  ;lower byte of address
mov h, 80h  ;higher byte of address
mov a, [hl] ;a <-- [h*256 + l]

CPU中存在许多寻址模式.因此，我们可以有一个不同的示例，例如只需一个寄存器和一个直接地址:

Many addressing modes exist in CPUs. So we can have a different example, e.g. with just a single register and an immediate address:

mov h, 80h
mov a, [2]  ;a <-- [h*256 + immediate]

它始终取决于特定的CPU体系结构.例如，Zilog Z80被称为8位CPU，但它也包含许多16位指令.您可以像这样对它进行索引寻址:

It always depends on a particular CPU architecture. For example Zilog Z80 is called 8-bit CPU but it also contains many 16-bit instructions. You can do indexed addressing on it like this:

mov ix, 8002h  ;base address of an array
mov a,[ix+20]  ;a <-- [ix + 20] i.e. read a byte from an array like ix[20] in C

注意: 那些旧的8位CPU使用8位累加器，即它们只能在8位寄存器中计算数学和其他算术运算，因此它们在软件计算级别上为8位.它们的内存访问单元是8位的，即它一次只能读取或写入一个字节的内存，因此它们在硬件级别上也是8位的.这些16位指令很慢，实际上它们是连续执行一对8位运算的.

Note: Those old 8-bit CPU's use an 8-bit accumulator, i.e. they can compute math and other arithmetic stuff only in an 8-bit register, so they are 8-bit on a software computation level. And their memory accessing unit is 8-bit, i.e. it can read or write just a single byte of memory at a time, so they are 8-bit on hardware level too. Those 16-bit instructions are slow, they actually do a pair of 8-bit operations in succession.

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