x86汇编牛顿平方根算法寄存器是1＃IND

Question

我正在尝试编写下面的C程序的汇编版本。它的意思是取9的平方根。一切似乎都在工作，直到第3次通过我的循环，然后当推到堆栈时，我和上下的值变为1＃IND。我不知道为什么，因为它正确地打印在标签 whileLoop 的正下方，但是第二次在标签 notAccurateEnough 之后直接打印时，我得到了错误的值，即使这两个标签之间的值不会发生变化，并且在之前的迭代中工作正常 下面是截图，显示它如何成功打印一次，然后无法在同一次迭代中第二次打印。

汇编代码：

.586
.MODEL FLAT
.STACK 4096                 
n=550
.data
    PUBLIC _sqrtX

upper REAL4 9.0
accuracy REAL4 0.00001
lower REAL4 1.0
guess REAL4 1.0
multValue REAL4 1.0
num REAL4 9.0
two DWORD 2

.code

_sqrtX  PROC

        finit   ;initialize fpu

        whileLoop:
        fld upper    ;push upper onto stack
        fld lower    ;push lower onto stack
        fsub         ;subtract upper-lower

        fld accuracy          ;push accuracy onto stack
        fcom                  ;compare accuracy with upper-lower

        fstsw ax              ;copy condition code bits to AX
        sahf                  ;shift condition code bits to flags
        jna notAccurateEnough ;jump if accuracy is greater
        jmp endnewt           ;end algorith, is less than accuracy

        notAccurateEnough:
        ;**************************
        ;************  Logic to implement:   guess = (lower + upper) / 2;
        ;**************************
        fld upper      ;push upper onto stack
        fld lower      ;push lower onto stack
        fadd           ;(lower + upper)
        fidiv two      ;divide by 2
        fstp guess     ;set guess to first stack element. guess = (lower + upper) / 2
        ;**************************

        ;**************************
        ;************  Logic to implement:  if((guess * guess) > number)
        ;**************************
        fld guess ;push guess onto stack as first element
        fld guess ;push guess onto stack again, it is now first and second element
        fmul      ;multiply first and second element
        fcom num  ;compare 9 to product of guess*guess:
        fstsw ax  ;copy condition code bits to AX
        sahf      ;shift condition code bits to flags
        jna SetLower ;if guess is less than num, call set lower and set lower = guess;
        ;*** else, guess > num
        fld guess ;push guess onto the stack
        fstp upper ;set upper equal to top of stack: upper = guess
        jmp whileLoop
        ;**************************

        setLower:
        fld guess   ;push guess onto stack
        fstp lower  ;store guess in lower
        jmp whileLoop

        endnewt:
        mov eax, 0
        ret

_sqrtX ENDP
END

C算法我试图在汇编中写：

double sqrtX (double x)
{
    const double ACCURACY=0.00001;
    double lower, upper, guess;
    double number;

    number = x;
    lower = 1;
    upper = number;

    while ((upper-lower) > ACCURACY)
    {
        guess = (lower + upper) / 2;
        if((guess * guess) > number)
        {
            upper = guess;
        }
        else
        {
            lower = guess;
        }
    }
    return (lower + upper)/2;
}

Answer 1

堆栈溢出。