我是NASM大会的新成员，我不知道为什么我的汇编代码会返回0.有人可以提供帮助或建议吗？该代码基于 C 函数模拟点积：
double dot(long int n, double x[], double y[])
{
double sum = 0.0;
long int i;
for(i=0; i<n; i++) sum = sum + x[i]*y[i];
return sum;
} // end dot
带值
long int n = 4;
double x[4] = {2.0, 3.0, 4.0, 5.0};
double y[4] = {5.0, -3.0, 2.0, 4.0};
double sum;
链接到nasm x86 64位汇编代码：
extern printf
section .bss
fltstorage: resq 1
tempflt: resq 1
section .data
dotMsg: db "dot n= ", 0
fmt_float: db "%f",10,0
fmt_dig: db "%ld",10, 0 ; print just one digit, e.g. 0 or 1
fmt_end: db 10, 0 ; just end line
sumOffset equ 32
section .text
global dot ;for dot function
dot:
push rbp ;push the stack
mov rbp, rsp ;callers stack
sub rsp, 8 ;make space for local variables (one quadword)
push rdi ;save registers
push rsi
;calculations go here
;RBX is off-limits
mov rsi, qword [rbp + 28] ;RSI now contains third argument
mov rdi, qword [rbp + 20] ;RDI now contains second argument
mov rcx, [rbp + 12] ;RCX should now contain first argument (this is a failure point- check it later)
mov r8, 0 ;clear results storage
dot_loop:
mov [fltstorage], rdi ;rdi is a pointer to element in x
fld qword [fltstorage] ;nth index of x array to st0
mov [fltstorage], rsi ;rsi is a pointer to element in y
fmul qword [fltstorage] ;multiply ST0 by element from y
add rdi, 8 ;increment rdi, go to next element in x
add rsi, 8 ;increment rsi, go to next element in y
mov [fltstorage], r8
fadd qword [fltstorage] ;add cumulative sum to fresh value
fstp qword [tempflt] ;move sum to storage
mov r8, tempflt ;reduce rcx
cmp rcx, 0 ;check if it's at the end of the loop
jg dot_loop ;if it is isn't,loop again
push rdi
push rsi
push rax
;;; print r8 to check output
mov rdi, fmt_float
mov rsi, r8
mov rax, 0
call printf
pop rax
pop rsi
pop rdi
mov rax, r8 ;prep result for return
add rsp, 8 ;balance out rsp from subtraction up top
pop rsi ;restore registers
pop rdi
mov rsp, rbp ;restore initial stack
pop rbp ;restore B register
ret