不带FPU的ASM数学运算

Question

我正在寻找一个例程，它可以帮助我在浮点数上做任何运算(它可以是加法)，但不使用浮点单元(FPU)。

我必须做一个程序，比较在没有FPU和有FPU的情况下操作的时间。第二部分，使用FPU已经完成，但我找不到任何好的例程。

Answer 1

不知道为什么人们都对它的复杂性感到震惊。加法、减法和乘法都不是很难。您将需要将数字拆分为符号、尾数和指数，然后使用整数数学来计算结果，最后将其放回浮点数。我碰巧有浮点乘法(64位GAS语法)的示例代码：

.globl main
main:
    subq $8, %rsp
    leaq format1(%rip), %rdi
    leaq (%rsp), %rsi
    leaq 4(%rsp), %rdx
    xor %eax, %eax
    call scanf

    movl (%rsp), %eax           # x
    movl 4(%rsp), %ecx          # y

    movl %eax, %edx             # edx = x
    xorl %ecx, %eax             # sgn(x*y) = sgn(x) ^ sgn(y)
    andl $0x80000000, %eax      # eax has correct sign bit now
    movl %ecx, %esi             # esi = y
    shrl $23, %esi
    andl $0xff, %esi            # esi = exp(y)
    movl %edx, %edi             # edi = x
    shrl $23, %edi
    andl $0xff, %edi            # edi = exp(x)
    addl %esi, %edi             # exp(x*y) = exp(x) + exp(y)
    subl $0x7f+23, %edi         # subtract bias and mantissa bits

    andl $0x007fffff, %edx      # edx = man(x)
    andl $0x007fffff, %ecx      # ecx = man(y)
    orl $0x00800000, %ecx       # put implicit leading 1 bit
    orl $0x00800000, %edx       # to both mantissas
    imulq %rcx, %rdx            # man(x*y) = man(x) * man(y)
    bsrq %rdx, %rcx             # find first set bit
    subl $23, %ecx              # leave 23 bits
    shrq %cl, %rdx              # shift the rest out
    andl $0x007fffff, %edx      # chop off implicit leading 1 bit
    addl %ecx, %edi             # adjust the exponent by the bits shifted out

    shll $23, %edi              # shift exponent into position
    orl %edi, %eax              # put exponent into result
    orl %edx, %eax              # put mantissa into result

    movss (%rsp), %xmm0
    movss 4(%rsp), %xmm1
    movd %eax, %xmm2
    cvtss2sd %xmm0, %xmm0
    cvtss2sd %xmm1, %xmm1
    cvtss2sd %xmm2, %xmm2
    movl $3, %eax
    leaq format2(%rip), %rdi
    call printf

    addq $8, %rsp
    ret
.data
    format1: .asciz "%f %f"
    format2: .asciz "%f * %f = %f\n"

对于zero、nan、inf、overflow之类的特殊情况，留给读者作为练习:)