如何在手臂编码即时价值? [英] How to encode immediate value in arm?

问题描述

假设有一个这样的实例:

Suppose there is an inst like this:

add  ip, ip, #0x5000

机器代码是

05 CA 8C E2

05 CA 8C E2

和

E2 8C CA 05 = 11100010100011001100 1010 00000101
imm = rotate_right(101B, 1010B*2) = 0x5000

但是，如果我们知道0x5000，我们怎么能得到101000000101?这种反向转换是一对一的对应吗?谢谢.

But if we know 0x5000, how can we get 101000000101? Is this reverse convert one-to-one correspondence? Thanks.

推荐答案

来自ARM ARM:

ADD 添加两个值.第一个值来自寄存器.第二个值可以是立即数，也可以是寄存器中的值，并且可以在加法之前进行移位.

ADD adds two values. The first value comes from a register. The second value can be either an immediate value or a value from a register, and can be shifted before the addition.

您看到的立即值已发生变化.指令的第11:0位是移位器操作数-在您的情况下为 0xA05 .

The immediate value you're seeing is being shifted. Bits 11:0 of your instruction are the shifter operand - in your case: 0xA05.

文档中稍后介绍了该寻址方式:

Later in the docs, that addressing mode is described:

< shifter_operand> 值是通过将8位立即数(向右)旋转到32位字中的任何偶数位位置而形成的.

The <shifter_operand> value is formed by rotating (to the right) an 8-bit immediate value to any even bit position in a 32-bit word.

因此，您特定的移位器操作数意味着将 0x05 向右旋转(2 * 10)位.

So your specific shifter operand means 0x05 rotated right by (2 * 10) bits.

如果要执行指令编码，则有几种选择.例如:

You have a few choices if you're doing the instruction encoding. For example:

0xA05 // rotate 0x05 right by 20
0xB14 // rotate 0x14 right by 22
0xC50 // rotate 0x50 right by 24

我手工编码了它们以进行分解:

I hand encoded them to disassemble:

$ xxd -r > example
00 05 CA 8C E2 14 CB 8C E2 50 CC 8C E2
$ arm-none-eabi-objdump -m arm -b binary -D example

example:     file format binary


Disassembly of section .data:

00000000 <.data>:
   0:   e28cca05    add ip, ip, #20480  ; 0x5000
   4:   e28ccb14    add ip, ip, #20480  ; 0x5000
   8:   e28ccc50    add ip, ip, #20480  ; 0x5000

这是一个可以找到编码的简单程序:

Here's a simple program that can find the encodings:

#include <stdio.h>
#include <stdlib.h>
#include <inttypes.h>

int main(int argc, char **argv)
{
    uint32_t encode = strtoul(argv[1], NULL, 0);
    int rotate;

    for (rotate = 0; rotate < 32; rotate += 2)
    {
        // print an encoding if the only significant bits 
        // fit into an 8-bit immediate
        if (!(encode & ~0xffU))
        {
            printf("0x%X%02X\n", rotate/2, encode);
        }

        // rotate left by two
        encode = (encode << 2) | (encode >> 30);
    }
    return 0;
}

并针对您的案例运行示例:

And an example run for your case:

$ ./example 0x5000
0xA05
0xB14
0xC50

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