Assembly 如何在汇编语言中找到奇数整数？
编写一个汇编语言程序，通过加倍和减半将两个整数相乘。下面是使用这种方法将A和B相乘的伪代码Assembly 如何在汇编语言中找到奇数整数？,assembly,Assembly,编写一个汇编语言程序，通过加倍和减半将两个整数相乘。下面是使用这种方法将A和B相乘的伪代码 Multiply A and B and store result in C: C = 0 ; Initialize to 0 while B is not equal to 0: if B is odd then C = C+A ; Add copy of A (even number left) A = 2*A ; Can be done quickly by shifting left B = B/
Multiply A and B and store result in C:
C = 0 ; Initialize to 0
while B is not equal to 0:
if B is odd then C = C+A ; Add copy of A (even number left)
A = 2*A ; Can be done quickly by shifting left
B = B/2 ; Can be done quickly by shifting right
我已经做了很多工作，如何使用shl测试奇数整数 它可以使用“
shrmov al, 0x10010010b
shr al, 1 ;CF = 0 even
mov al, 0x10010011b
shr al, 1 ;CF = 1 odd
jnb __lable ;jump if CF = 0
or
jb __lable ;jump if CF = 1
对于x86_64，您可以使用
shr segment .text
mulByBits:
; input: 2x uint32_t in edi, esi (System V AMD64 ABI calling convention)
; output: uint64_t in rax
; modifies also: rcx
xor eax, eax ; rax = 0 "C"
mov edi, edi ; clear upper 32b of input "A" (extend to 64b)
.mulLoop:
lea rcx, [rax + rdi] ; rcx = C + A (new C in case B is odd)
add rdi, rdi ; A *= 2 (for next loop)
shr esi, 1 ; B >>= 1 (sets ZF and CF)
cmovc rax, rcx ; if B was odd, update the sum to new C
jnz .mulLoop ; repeat until B is zero
ret
global _start
_start: ; run some hardcoded simple tests, verify in debugger
mov edi, 143254 ; "normal" values test
mov esi, 43526
call mulByBits
mov rbx, 6235273604 ; expected result, compare with rax
mov edi, 0
mov esi, 0
call mulByBits
mov rbx, 0
mov edi, 43257432
mov esi, 0
call mulByBits
mov rbx, 0
mov edi, 0
mov esi, 432543
call mulByBits
mov rbx, 0
mov edi, 3276547234
mov esi, 1
call mulByBits
mov rbx, 3276547234
mov edi, 1
mov esi, 3276547234
call mulByBits
mov rbx, 3276547234
mov edi, ~0 ; UINT_MAX * UINT_MAX
mov esi, ~0
call mulByBits
mov rbx, 0xFFFFFFFE00000001
mov rdi, 0xE00000004 ; garbage in upper 32 bits of inputs
mov rsi, 0xE00000004 ; should be multiplied as 4*4
call mulByBits
mov rbx, 0x10
; exit back to linux
mov eax, 60
xor edi, edi
syscall
在大多数CPU上，加法是最有效的左移1的方法<代码>添加相同的，相同的可以在比
shl rdi，1shrjcjnc添加rdi，rdishl rdi，1adc-same，与rcl-by-1类似）。不幸的是，现代Intel CPU没有将左移位1的短格式操作码解码/运行到任何ALU端口上运行的add-like uop中。@PeterCordes确实，谢谢，修改了源代码，因为这应该是为了炫耀“正确的”x86_64，而不是按照字面上的赋值。（虽然mul
本身确实存在，并且比这个比特碰撞更快，所以它仍然只是为了锻炼而锻炼）添加相同的，相同的是左移，并且shl
乘以1并不完全慢；但大多数CPU最多有2个整数移位端口。不管怎样，看起来不错，我明天会投票。我今天的投票已经用光了。顺便说一句，mov-edx，edi
在消除mov的英特尔CPU上的延迟比mov-edi，edi
低1个周期。(). 但是我认为为你的函数重写寄存器分配是不值得的。是的，值多少取决于数据分布。在任何工作完成之前，分支未命中意味着您回到了dep链的开始，而如果以后的指令主要依赖于乘法结果，那么在~log2（n）次迭代后离开循环的预测失误也不是什么大问题。（单个循环shr
dep链可以在循环的其余部分之前运行。）但如果最高设置位经常有很大的不同，则可以节省大量迭代。使用可变计数移位和tzcnt
仅在设定位上循环也是可能的，并且可能在设定位较少的情况下获胜。
segment .text
mulByBits:
; input: 2x uint32_t in edi, esi (System V AMD64 ABI calling convention)
; output: uint64_t in rax
; modifies also: rcx
xor eax, eax ; rax = 0 "C"
mov edi, edi ; clear upper 32b of input "A" (extend to 64b)
.mulLoop:
lea rcx, [rax + rdi] ; rcx = C + A (new C in case B is odd)
add rdi, rdi ; A *= 2 (for next loop)
shr esi, 1 ; B >>= 1 (sets ZF and CF)
cmovc rax, rcx ; if B was odd, update the sum to new C
jnz .mulLoop ; repeat until B is zero
ret
global _start
_start: ; run some hardcoded simple tests, verify in debugger
mov edi, 143254 ; "normal" values test
mov esi, 43526
call mulByBits
mov rbx, 6235273604 ; expected result, compare with rax
mov edi, 0
mov esi, 0
call mulByBits
mov rbx, 0
mov edi, 43257432
mov esi, 0
call mulByBits
mov rbx, 0
mov edi, 0
mov esi, 432543
call mulByBits
mov rbx, 0
mov edi, 3276547234
mov esi, 1
call mulByBits
mov rbx, 3276547234
mov edi, 1
mov esi, 3276547234
call mulByBits
mov rbx, 3276547234
mov edi, ~0 ; UINT_MAX * UINT_MAX
mov esi, ~0
call mulByBits
mov rbx, 0xFFFFFFFE00000001
mov rdi, 0xE00000004 ; garbage in upper 32 bits of inputs
mov rsi, 0xE00000004 ; should be multiplied as 4*4
call mulByBits
mov rbx, 0x10
; exit back to linux
mov eax, 60
xor edi, edi
syscall