什么时候应该在x86中使用size指令? [英] When should I use size directives in x86?

问题描述

何时在x86中使用size指令似乎有点模棱两可. 此x86组装指南表示以下内容:

When to use size directives in x86 seems a bit ambiguous. This x86 assembly guide says the following:

通常，给定内存中数据项的预期大小 可以从其中的汇编代码指令中推断出该地址 参考.例如，在以上所有说明中， 可以从寄存器的大小推断出存储区域 操作数.当我们加载32位寄存器时，汇编器可以 推断我们所指的内存区域为4个字节宽. 当我们将一个字节寄存器的值存储到内存中时， 汇编程序可以推断出我们希望该地址引用单个地址 内存中的字节.

In general, the intended size of the of the data item at a given memory address can be inferred from the assembly code instruction in which it is referenced. For example, in all of the above instructions, the size of the memory regions could be inferred from the size of the register operand. When we were loading a 32-bit register, the assembler could infer that the region of memory we were referring to was 4 bytes wide. When we were storing the value of a one byte register to memory, the assembler could infer that we wanted the address to refer to a single byte in memory.

他们给出的例子非常简单，例如将立即数移到寄存器中.
但是，如何处理更复杂的情况，例如:

The examples they give are pretty trivial, such as mov'ing an immediate value into a register.
But what about more complex situations, such as the following:

mov    QWORD PTR [rip+0x21b520], 0x1

在这种情况下，QWORD PTR size指令不是多余的，因为根据上述指南，由于RIP为8字节，因此可以假定我们要将8字节移到目标寄存器中? x86体系结构上的size指令的最终规则是什么?谢谢，我在任何地方都找不到答案.

In this case, isn't the QWORD PTR size directive redundant since, according to the above guide, it can be assumed that we want to move 8 bytes into the destination register due to the fact that RIP is 8 bytes? What are the definitive rules for size directives on the x86 architecture? I couldn't find an answer for this anywhere, thanks.

更新:正如Ross指出的那样，以上示例中的目的地不是寄存器.这是一个更相关的示例:

Update: As Ross pointed out, the destination in the above example isn't a register. Here's a more relevant example:

mov    esi, DWORD PTR [rax*4+0x419260] 

在这种情况下，是否不能因为ESI为4字节而假定我们要移动4个字节，从而使DWORD PTR指令冗余?

In this case, can't it be assumed that we want to move 4 bytes because ESI is 4 bytes, making the DWORD PTR directive redundant?

推荐答案

您是对的；这是相当模棱两可的.假设我们在谈论Intel语法，的确可以使您经常使用size指令而不使用 not .只要汇编程序能够自动找出答案，它们都是可选的.例如，在指令中

You're right; it is rather ambiguous. Assuming we're talking about Intel syntax, it is true that you can often get away with not using size directives. Any time the assembler can figure it out automatically, they are optional. For example, in the instruction

mov    esi, DWORD PTR [rax*4+0x419260] 

DWORD PTR说明符是可选的，这正是您所想的原因:汇编程序可以确定它是要移动DWORD大小的值，因为该值已被移动到DWORD大小的寄存器中.

the DWORD PTR specifier is optional for exactly the reason you suppose: the assembler can figure out that it is to move a DWORD-sized value, since the value is being moved into a DWORD-sized register.

类似地，在

mov    rsi, QWORD PTR [rax*4+0x419260] 

出于完全相同的原因，QWORD PTR说明符是可选的.

the QWORD PTR specifier is optional for the exact same reason.

但是它并不总是可选的.考虑您的第一个示例:

But it is not always optional. Consider your first example:

mov    QWORD PTR [rip+0x21b520], 0x1

在这里，QWORD PTR说明符是不是可选的.没有它，汇编器将不知道要从地址rip+0x21b520开始存储的大小值. 0x1应该存储为BYTE吗?扩展到WORD吗?双字?一个QWORD?一些汇编程序可能会猜测，但是如果不明确指定所需的内容，您将无法保证正确的结果.

Here, the QWORD PTR specifier is not optional. Without it, the assembler has no idea what size value you want to store starting at the address rip+0x21b520. Should 0x1 be stored as a BYTE? Extended to a WORD? A DWORD? A QWORD? Some assemblers might guess, but you can't be assured of the correct result without explicitly specifying what you want.

换句话说，当值位于 register 操作数中时，大小说明符是可选的，因为汇编程序可以根据寄存器的大小确定大小.但是，如果要处理立即数或内存操作数，则可能需要使用大小说明符，以确保获得所需的结果.

In other words, when the value is in a register operand, the size specifier is optional because the assembler can figure out the size based on the size of the register. However, if you're dealing with an immediate value or a memory operand, the size specifier is probably required to ensure you get the results you want.

就我个人而言，我更喜欢在编写代码时总是 包含大小.它是几个字符，但要输入，但它迫使我考虑一下并明确说明我想要的内容.如果我搞砸了并且编写了一个不匹配的代码，那么汇编器将大声地向我尖叫，这不只一次捕获了错误.我也认为在此安装可增强可读性.所以在这里，我同意 old_timer ，尽管他的观点似乎并不受欢迎.

Personally, I prefer to always include the size when I write code. It's a couple of characters more typing, but it forces me to think about it and state explicitly what I want. If I screw up and code a mismatch, then the assembler will scream loudly at me, which has caught bugs more than once. I also think having it there enhances readability. So here I agree with old_timer, even though his perspective appears to be somewhat unpopular.

反汇编程序的输出也往往很冗长，包括大小说明符，即使它们是可选的.汉斯·帕森特(Hans Passant)在评论中提出，这是为了保持与一直需要它们的老式组装商的向后兼容性，但我不确定这是真的.它可能是其中的一部分，但是根据我的经验，反汇编程序在很多上往往会用很多不同的方式出现，因此我认为这只是为了简化分析陌生代码的过程.

Disassemblers also tend to be verbose in their outputs, including the size specifiers even when they are optional. Hans Passant theorized in the comments this was to preserve backwards-compatibility with old-school assemblers that always needed these, but I'm not sure that's true. It might be part of it, but in my experience, disassemblers tend to be wordy in lots of different ways, and I think this is just to make it easier to analyze code with which you are unfamiliar.

请注意，AT& T语法使用了稍微不同的方式.它没有将大小写为操作数的前缀，而是为指令助记符添加了后缀:b表示字节，w表示字，l表示dword，q表示qword.因此，前面的三个示例变为:

Note that AT&T syntax uses a slightly different tact. Rather than writing the size as a prefix to the operand, it adds a suffix to the instruction mnemonic: b for byte, w for word, l for dword, and q for qword. So, the three previous examples become:

movl    0x419260(,%rax,4), %esi
movq    0x419260(,%rax,4), %rsi
movq    $0x1, 0x21b520(%rip)

同样，在前两个指令中，l和q前缀是可选的，因为汇编程序可以推断出适当的大小.在最后一条指令上，就像在Intel语法中一样，前缀是非可选的.因此，AT& T语法中的内容与Intel语法相同，只是大小说明符的格式不同.

Again, on the first two instructions, the l and q prefixes are optional, because the assembler can deduce the appropriate size. On the last instruction, just like in Intel syntax, the prefix is non-optional. So, the same thing in AT&T syntax as Intel syntax, just a different format for the size specifiers.

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