ARMv6汇编中= label(等号)和[label](括号)之间有什么区别? [英] What is the difference between =label (equals sign) and [label] (brackets) in ARMv6 assembly?
问题描述
我正在跟烘焙Pi 剑桥大学的课程，其中针对Raspberry Pi的ARMv6指令集构建了一个简单的操作系统.
I'm following along with the Baking Pi course from Cambridge University, in which a simple operating system is built in the ARMv6 instruction set, targeting the Raspberry Pi.
到目前为止，我们一直在使用两种通过ldr指令将数据加载到寄存器中的方式，现在我意识到我正在一起使用它们，但我还不完全了解它们的作用.
We've been using two ways of loading data into registers via the ldr instruction so far and I realize now that I'm using them together, I don't fully understand what they both do.
因此，我使用了ldr r0,=0x20200000之类的东西，我实际上将其理解为将存储在内存位置0x20200000中的数据读入寄存器r0.
So I've used things like ldr r0,=0x20200000, which I actually understood as "read the data stored at the memory location 0x20200000 into register r0.
然后我使用了类似的东西:
Then I've used things like:
ldr r0,[r1,#4]
据我所知，这是将存储在r1指向的内存地址上的数据以4个字节的偏移量读入寄存器r0".
Which I've understood as being "read the data stored at the memory address pointed to by r1, at an offset of 4 bytes, into register r0".
然后我遇到了这个
ldr r0,=pattern
ldr r0,[r0]
pattern在这里是.data部分中的.int(位图表示LED的开/关状态序列).读完此书后，我意识到我以前对=foo的理解一定是错误的，否则以上两个说明都将做同样的事情.
pattern here is a .int in the .data section (a bitmap representing a sequence of on/off states for an LED). I realize upon reading this, that my previous understanding of =foo must be wrong, otherwise both of the above instructions would do the same thing.
=x语法基本上更像C中的指针，而[x]语法好像实际上是读取了x所指向的内存吗?
Is the =x syntax basically more like a pointer in C, while the [x] syntax is as if the memory that is being pointed to by x is actually read?
假设下面的C语言中的ptr是int*，那么我对等效组装(从概念上讲，从字面上看)的思考是否有意义?
Let's say ptr in the C below is an int*, do my comments thinking about equivalent assembly (conceptually, not literally) make any sense?
r0 = ptr; /* equivalent to: ldr r0,=ptr */
r0 = *ptr; /* equivalent to: ldr r0,[ptr] */
r0 = *(ptr+4) /* equivalent to: ldr r0,[ptr,#4] */
推荐答案
ldr r0,=something
...
something:
表示将标签的地址加载到寄存器r0中.然后，汇编程序在ldr指令可及的位置添加一个单词，并将其替换为
means load the address of the label something into the register r0. The assembler then adds a word somewhere in reach of the ldr instruction and replaces it with a
ldr r0,[pc,#offset]
说明
所以这个快捷方式
ldr r0,=0x12345678
表示将0x12345678加载到r0中.
means load 0x12345678 into r0.
由于是固定长度的指令，您不能在一条指令中将完整的32位立即数加载到寄存器中，它可能需要许多指令才能完全加载32位数字的寄存器.在很大程度上取决于数量.例如
being mostly fixed length instructions, you cant load a full 32 bit immediate into a register in one instruction, it can take a number of instructions to completely load a register with a 32 bit number. Depends heavily on the number. For example
ldr r0,=0x00010000
如果是ARM指令，则将用单个指令mov r0，#0x00010000替换为gnu汇编程序；对于拇指指令，尽管它仍必须是ldr r0，[pc，#offset]
will get replaced by the gnu assembler with a single instruction mov r0,#0x00010000 if it is an ARM instruction, for a thumb instruction though it may still have to be ldr r0,[pc,#offset]
这些ldr rd，=事物是快捷方式，伪指令，不是真实的.
These ldr rd,=things are a shortcut, pseudo instructions, not real.
ldr rd,[rm,#offset]
ldr rd,[rm,rn]
是真实的指令，是指从存储器的地址rm + offset或rm + rn读取，并将读取的值放入寄存器rd
are real instructions and mean read from memory at address rm+offset or rm+rn and take the value read and put it in the register rd
= something更类似于C中的& something
the =something is more like &something in C.
unsigned int something;
unsigned int r0;
unsigned int r1;
r0 = &something;
r1 = *(unsigned int *)r0;
并在组装中
something:
.word 0
ldr r0,=something
ldr r1,[r0]
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