Linux IMUL指令寄存器

我写了这个简单的程序来计算递归序列的第I个元素。序列基本上看起来像

a(n)=a(n-1)*a(n-2)

前两个元素是-1和-3。我使用

imul

进行乘法，并且由于我在网络中的发现，我应该能够使用我想要的任何寄存器，但是program为第三个元素返回0。切换到

add

时，它会按预期工作。 下面是我递归调用函数并进行乘法的片段（如图所示，我使用堆栈存储变量）

基本上，问题是“为什么它不起作用”：p

注：如果需要完整代码：

.data
STDOUT = 1
SYSWRITE = 1
HOW_MANY = 3                # which number to calculate
SYSEXIT = 60
EXIT_SUCCESS = 0
FIRST = -1                  # first element of the sequence
SECOND = -3                 # second element of the sequence
NUMBER_BEGIN = 0x30
OUTPUT_BASE = 10
NEW_LINE = '\n'
PLUS = '+'
MINUS = '-'

.bss
.comm textin, 512
.comm textout, 512
.comm text2, 512
.comm znak, 1

.text
.globl _start

_start:

#
# Calling function to calculate ith element
#
mov $HOW_MANY, %r8
sub $1, %r8
push %r8                    # push r8 (function argument) to stack
call calculate              # call function to calculate
add $8, %rsp                # removing parameter from stack

# now we should've have result in rbx  
#
mov $0, %r15 # Flaga znaku (domyślnie 0 = +)
cmp $0, %rbx
jge to_ascii # Pomiń jeśli liczba jest dodatnia
not %rbx # Odwrócenie bitów liczby i dodanie 1,
inc %rbx # aby uzyskać jej wartość bezwzględną.
mov $1, %r15 # Ustawienie flagi znaku na 1 = -.

to_ascii:
mov %rbx, %rax # result goes to rax
mov $OUTPUT_BASE, %rbx
mov $0, %rcx
jmp loop

loop:
mov $0, %rdx
div %rbx # divide rax by rbx, rest in rdx
add $NUMBER_BEGIN, %rdx # rest in rdx is a next position number
mov %dl, text2(, %rcx, 1)

inc %rcx
cmp $0, %rax
jne loop
jmp inverse

inverse:
mov $0, %rdi
mov %rcx, %rsi
dec %rsi
jmp inversev2

inversev2:
mov text2(, %rsi, 1), %rax
mov %rax, textout(, %rdi, 1)
inc %rdi
dec %rsi
cmp %rcx, %rdi
jle inversev2

push %rcx # legth of the answer goes to stack

mov $0, %r10 # want sign at the first position
movb $PLUS, znak(, %r10, 1)

cmp $0, %r15 # r15 register contains info about the sign
je next # 0 = +, so nothing has to be done

movb $MINUS, znak(, %r10, 1) # otherwise set it to minus

next:   # show sign
mov $SYSWRITE, %rax
mov $STDOUT, %rdi
mov $znak, %rsi
mov $1, %rdx
syscall

pop %rcx

movb $NEW_LINE, textout(, %rcx, 1)
inc %rcx

mov $SYSWRITE, %rax
mov $STDOUT, %rdi
mov $textout, %rsi
mov %rcx, %rdx
syscall

mov $SYSEXIT, %rax
mov $EXIT_SUCCESS, %rdi
syscall

# recursive function calculating ith element of a given sequence
# sequence =
# n_1 = -1
# n_2 = -3
# n_i = n_(i-1)*n_(i-2)
calculate:
push %rbp                   # push rbp to stack to save it's value
mov %rsp, %rbp              # now stack pointer is stored in rbp

sub $8, %rsp
mov 16(%rbp), %rax

cmp $1, %rax
jl first
je second

mov $0, %rcx

# wywołanie dla n_(i-1)
dec %rax

push %rcx
push %rax
call calculate
pop %rax
pop %rcx # przepisać na rejestry imula
imul %rcx, %rbx

# wywołanie dla n_(i-2)
dec %rax
push %rcx
push %rax
call calculate
pop %rax
pop %rcx
imul %rcx, %rbx

return:
mov %rcx, %rbx
mov %rbp, %rsp
pop %rbp
ret

first:
mov $FIRST, %rbx
mov %rbp, %rsp
pop %rbp
ret

second:
mov $SECOND, %rbx
mov %rbp, %rsp
pop %rbp
ret

---

您正在将%rcx设定为零，然后再乘以该值，因此您将始终具有零的乘积

也许你想改变

mov $0, %rcx

到

我认为你还需要扭转局面

imul %rcx, %rbx

到

---

（我不熟悉汇编程序的那种风格）

请在问题中内联发布完整的代码。好了。我想如果我添加到pastebin Insted的链接，问题会变得更短。是的，我已经切换了，因为我希望我的结果是

%rcx

而不是

%rbx

mov $1, %rcx

imul %rcx, %rbx

imul %rbx, %rcx