Suppose there is an inst like this:
add ip, ip, #0x5000
the machine code is
05 CA 8C E2
and
E2 8C CA 05 = 11100010100011001100 1010 00000101
imm = rotate_right(101B, 1010B*2) = 0x5000
But if we know 0x5000, how can we get 101000000101? Is this reverse convert one-to-one correspondence? Thanks.
From the ARM ARM:
ADDadds 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.
The immediate value you're seeing is being shifted. Bits 11:0 of your instruction are the shifter operand - in your case: 0xA05.
Later in the docs, that addressing mode is described:
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.
So your specific shifter operand means 0x05 rotated right by (2 * 10) bits.
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
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
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;
}
And an example run for your case:
$ ./example 0x5000
0xA05
0xB14
0xC50
