ARM v8 中的 LDUR 和 STUR [英] LDUR and STUR in ARM v8

问题描述

我上过几门涉及 ARMv8 汇编的课程，但是两位老师都以不同的方式描述了 LDUR/STUR 指令，现在我已经迷失了方向.有人可以帮忙澄清一下吗?

I've had a couple of courses that touched on ARMv8 assembly, but both teachers described LDUR/STUR instructions a different way and now I've become pretty lost. Can someone help to clarify?

如果我有指令:

LDUR R3, [R1, #8]

我会将答案放在 R3 中，但是我从 R1 中得到了什么以及偏移量如何运作?这就像一个逻辑转变吗?ARM 手册将其描述为字节偏移".但随后没有描述该偏移量如何在 R1 上发挥作用.我是否移动了存储在 R1 中的值(比如 R1 的值是 50)，还是我需要考虑 R1 之外的内存地址?其他消息来源说我需要以某种方式将 R1 视为一个数组?

I'll be putting the answer in R3, but what am I taking from R1 and how does the offset operate? Is it like a logical shift? The ARM manual describes it as "byte offset" but then doesn't describe how that offset functions on R1. Do I shift the value stored in R1 (say R1 has value 50 in it) or is there a memory address outside of the R1 that I need to be thinking about? Other sources say I need to think of R1 as an array somehow?

推荐答案

LDUR 是 Load (unscaled) Register.它从一个地址加上一个寄存器的偏移量加载一个值(32 位或 64 位).unscaled 意味着在机器代码中，偏移量将不用像 ldr 使用的缩放偏移量编码，即不会应用移位到立即偏移位.偏移量(simm 有符号立即数)将被添加到基址寄存器 Xn|SP.

LDUR is Load (unscaled) Register. It loads a value (32-bits or 64-bits) from an address plus an offset to a register. unscaled means that in the machine-code, the offset will not be encoded with a scaled offset like ldr uses, i.e. no shift will be applied to the immediate offset bits. The offset (simm signed immediate) will be added to the base register Xn|SP.

因此，与 ldr

这些是 LDUR 的原型:

These are the prototypes for LDUR:

    -- loads a 32-bit value
    LDUR <Wt>, [<Xn|SP>{, #<simm>}]

    -- loads a 64-bit value
    LDUR <Xt>, [<Xn|SP>{, #<simm>}]

STUR 是存储(未缩放)寄存器，工作方式相同，但它将寄存器中的值存储到内存中.

STUR is Store (unscaled) Register and works in the same way but it stores the value in a register to memory.

这些是 STUR 的原型:

These are the prototypes for STUR:

    -- stores a 32-bit register
    STUR <Wt>, [<Xn|SP>{, #<simm>}]

    -- stores a 64-bit register
    STUR <Xt>, [<Xn|SP>{, #<simm>}]

LDUR/STUR 允许访问与操作数的大小不对齐的 32/64 位值.例如，存储在地址 0x52 处的 32 位值.

LDUR/STUR allow accessing 32/64-bit values when they are not aligned to the size of the operand. For example, a 32-bit value stored at address 0x52.

在你的例子中，

    LDUR R3, [R1, #8]

这条指令会将R1指向的值加上8个字节加载到R3.这就是 ARM 参考手册中 byte offset 的意思.因此，如果 R1 持有值 0x50，这将加载存储在地址 0x58 的值.R1 的值不会被修改.

this instruction will load to R3 the value pointed by R1 plus 8 bytes. This is what the ARM Reference Manual means by byte offset. So if R1 holds the value 0x50, this will load the value stored at address 0x58. The value of R1 will not be modified.

指令 LDR R3, [R1, #8](LDR (immediate) Unsigned offset 变体)产生相同的操作，然而，原型是不同的:

The instruction LDR R3, [R1, #8] (LDR (immediate) the Unsigned offset variant) produces the same operation, however, the prototype is different:

-- loads a 32-bit value
LDR <Wt>, [<Xn|SP>{, #<pimm>}]

-- loads a 64-bit value
LDR <Xt>, [<Xn|SP>{, #<pimm>}]

直接偏移pimm不同，LDUR使用simm.这意味着以不同的方式解释偏移量.第一个 (pimm) 是一个正偏移量，其范围对于 32 位变体和 64 位变体是不同的.

The immediate offset pimm is different, LDUR uses a simm. This means that the offset is interpreted in a different way. The first (pimm) is a positive offset and its range is different for the 32-bit variant and the 64-bit variant.

在 32 位版本中:

• 范围从0到16380，只能是4的倍数

在 64 位版本中:

• 范围从 0 到 32760，并且只能是 8 的倍数

这意味着 LDUR 和 LDR(立即)的一些偏移组合将产生相同的操作.

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