ARM:为什么立即数只有 12 位? [英] ARM: Why only 12 bits for immediate constants?

问题描述

这是什么意思:我只有 12 位用于立即常数，所以我可以只表示从 0 到 2^12 = 4096 的立即常数吗?操作数 2 如果是寄存器，可以有 32 位，但为什么立即数只有 12 位呢?这个数字是从哪里来的?

what does it mean : I have only 12 bits for immediate constants, so can I represent immediate constants only from 0 to 2^12 = 4096 ? Operand 2 , if it's a register, can have 32 bits, but why only 12 bits for immediate constants? Where does this number come from?

推荐答案

由指令集定义.例如.MOV 指令编码为

It defined by the instruction set. E.g. the MOV instruction is encoded as

31 28 | 27 26 | 25 | 24 23 22 21 20 | 19   16 | 15    12 | 11        0      |
cond  | 0  0  | I  | 1  1  0  1  S  | SBZ     | Rd       | shifter operand  |

(see "ARM Architecture Reference Manual, 4.1.29 "MOV")

立即"常量在移位器操作数"中编码，只有 12 位.其他指令有类似的定义有时是其他宽度.

"Immediate" constants are encoded in the "shifter operand" which is 12 bits only. Other instructions have similar definitions are sometimes other widths.

存在此限制是因为 - 与 x86 不同 - 使用 Thumb(2) 时，ARM 上的指令始终为 32 位或有时为 16 位.为了支持 12 位二进制数字无法直接表示的值，移位器操作数允许不同的寻址模式(例如左移、右移、旋转).

This limitation exists because -- unlike on x86 -- instructions on ARM are always 32 bit or sometimes 16 bit when using Thumb(2). To support values which can not be expressed directly by a 12 bit binary digit, the shifter operand allows different addressing modes (e.g. left-shift, right-shift, rotating).

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