assembly - 如何在arm中编码立即值？

Question

假设有一个这样的实例:

add  ip, ip, #0x5000

机器码是

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，我们怎样才能得到101000000101呢？这个反向转换是一一对应的吗？谢谢。

Answer 1

来自 ARM ARM:

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.

您所看到的直接值(value)正在发生变化。指令的位 11:0 是移位器操作数 - 在您的情况下:0xA05 .

文档后面描述了该寻址模式:

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.

所以你的特定移位操作数意味着 0x05向右旋转 (2 * 10)位。

如果您正在进行指令编码，您有几种选择。例如:

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

我手工编码它们以进行反汇编:

$ 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:

#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;
}

以及针对您的案例运行的示例:

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