为什么必须取消引用数据标签才能在其中存储内容:Assembly 8086 FASM [英] Why do you have to dereference the label of data to store something in there: Assembly 8086 FASM

问题描述

考虑以下示例:

mov al, [variable1]

这是有道理的，因为您在寄存器 al 中保存了变量 1 的内容(因此取消了对它的引用).

This makes sense, as you save in the register al the content (thus dereferencing it) of variable1.

现在考虑这个例子:

mov dword L6, 1

这是不正确的；正确的形式是

This isn't correct; the correct form is

mov dword [L6], 1

如果我们将 1 移动到内存中，我应该告诉它存储在那个地址而不是内容中.假设您要告诉某人去某个地方参加派对，您需要向他们提供派对的地址，而不是那里的人.

If we're moving 1 to memory, I should tell it to be stored at that address and not at the content. Say you're telling someone to go to a certain place for a party, you'd give them the address of the party and not the people in there.

那么，这是为什么呢?

推荐答案

x86 没有内存间接数据寻址.这是 0 和 1 级间接与作为立即数的符号地址之间的区别，而不是 1 与 2.

x86 doesn't have memory-indirect data addressing. It's the difference between 0 and 1 levels of indirection vs. the symbol address as an immediate, not 1 vs. 2.

与 C 不同，在 FASM(和 NASM)语法中，提及裸符号名称是地址作为值.

Unlike C, in FASM (and NASM) syntax, mentioning a bare symbol name is the address as a value.

例如mov eax, symbol 设置 EAX = 放置该标签的地址，使用 mov eax, imm32 指令，而不是从内存中加载.

e.g. mov eax, symbol sets EAX = the address where you put that label, with a mov eax, imm32 instruction, not a load from memory.

当然mov符号，eax不能工作，因为立即数不能作为目标，只有[disp32]寻址模式.

Of course mov symbol, eax can't work because an immediate constant can't be a destination, only a [disp32] addressing mode.

如果我将 mov 视为 write ，它会更有意义.我可以将 mov 视为写入而不是移动"吗?

if I think of mov as write it makes a lot more sense. Can I think of mov as write instead of "moving?

是的.mov 只是从源操作数复制到目标操作数.这就是目标必须指向内存或寄存器的原因.但源操作数可以是立即数、寄存器或内存.

Yes. mov just copies from the source operand to the destination operand. That's why the destination has to be pointed-to memory or a register. But the source operand can be an immediate, register, or memory.

(当然，对于单个 mov 指令，您不能同时拥有 src 和 dst 内存；有 不同形式用于不同类型的操作数.)

(Of course you can't have both src and dst be memory for a single mov instruction; there are different forms for different kinds of operands.)

FASM 和 NASM 中的 asm 符号有点像 C char arr[].

你不能做 arr = x; - asm 等价物是 mov arr, al 这也不合法(在 FASM 和 NASM 语法中).

You can't do arr = x; - the asm equivalent is mov arr, al which isn't legal either (in FASM and NASM syntax).

但是你可以做 *arr = x; 在 asm 中是 mov [arr], al

But you can do *arr = x; which in asm is mov [arr], al

仅供参考:MASM 语法不同:mov sym, al 不是在所有上下文中用作数字占位符的符号，而是与 mov [sym], al 和括号是可选的，没有寄存器没有意义.

FYI: MASM syntax is different: instead of symbols working as a placeholder for a number in all contexts, mov sym, al is a store the same as mov [sym], al and the brackets are optional and meaningless without a register.

如果您看到任何 MASM 示例或 GCC/clang .intel_syntax noprefix 输出；这就是正在发生的事情.

If you see any MASM examples or GCC/clang .intel_syntax noprefix output; that's what's going on.

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