Arm64- why all filled with 1 the lower 32bit? Isn't it should be only 1 value?

Question

mov x0, #1000

mov w6, #-1
add x7,x0,x6, sxtw

what will be the value of x6 and x7? [In hex] Here ans x6=0xffffffff //why all value of lower half filled with 1? and how sxtw work when the ans will be 999 in decimal. I want to know details how this happening.

But, do you know about two's complement arithmetic? As 32 bit numbers, `-1` and `0xffffffff` are the same. — Nate Eldredge, Aug 27 '22 at 05:56
The idea is that the bit pattern `0xffffffff` can be used to represent the integer -1, because adding `0xffffffff` to any 32-bit number (and ignoring any carry) has exactly the same mathematical effect as adding -1 (i.e. it subtracts 1). Try some examples. — Nate Eldredge, Aug 27 '22 at 06:00
But anyway, the comments section of this question is not the place for me to teach you about two's complement representations. It is a very standard and fundamental concept, and you ought to read about it in a standard textbook. — Nate Eldredge, Aug 27 '22 at 06:02
Thanks I got it now but 3rd statement code i did not found in book mean sxtw — Parvez Hossain, Aug 27 '22 at 06:03
You can look up the instruction in any instruction set reference, such as the official Architecture Reference Manual. It takes the low 32 bits of x6, [sign extends](https://en.wikipedia.org/wiki/Sign_extension) to a 64-bit value, adds that value to x0, and places the result into x7. — Nate Eldredge, Aug 27 '22 at 06:06
Thanks a lot. I got it all. Should i delete my question now? — Parvez Hossain, Aug 27 '22 at 06:07

score 0 · Answer 1 · answered Aug 27 '22 at 06:14

// 1= 000000000000000000000001 (On 32bit)
// -1= 111111111111111111111111
// So -1 in hex = 0xffffffff

The lower half of x6 will be filled with above value.

On 3rd statement it sign extends the value to 64bit so -1 is correct then it add 1000+(-1) and become value 999

Arm64- why all filled with 1 the lower 32bit? Isn't it should be only 1 value?

1 Answers1