c - 在汇编中实现返回最大数和最小数之差的 C 函数

Question

我是汇编初学者，我正在尝试编写一个汇编函数来实现这段 C 代码中的 largestdif 函数(它接收 3 个整数作为参数)。它基本上需要找到 3 个参数的最小值和最大值，然后返回最大值和最小值之间的减法:

#include <stdio.h>
extern int largestdif( int n1, int n2, int n3 );
int main( ) {
printf( "largestdif: %d\n", largestdif( 100, 30, 10 ) );
printf( "largestdif: %d\n", largestdif( 0, -1, 1 ) );
return 0;
}

这是我在汇编中的一段代码，看起来有点困惑，但我希望你能告诉我这有什么问题，因为它出现了段错误。我真的不知道还有什么可做的，这一定与寄存器没有被足够调用有关。似乎无法掌握答案，所以是的，我希望有人能帮助我。提前致谢。

.global largestdif
.text
largestdif:
push %ebp
mov %esp, %ebp
mov 8(%ebp), %ebx
push %ebx
mov 12(%ebp), %ecx
push %ecx
mov 16(%ebp), %edx
push %edx

#compare n1 and n2 to find the largest number
cmp %edx, %ecx
jl n1_menor_n2
    #block else
    cmp %edx, %ebx #compare first and third numbers
    jl firstlower
        #bloco else
        mov %edx, %eax #already have largest number
        jmp continue
    firstlower:
        mov %ebx, %eax #already have largest number
        jmp continue
n1_menor_n2:
    cmp %ecx, %ebx #compare second and third numbers
    jl secondlower
        #block else
        mov %ecx, %eax # already have largest
        jmp continue
    secondlower:
        mov %ebx, %eax # already have largest
        jmp continue


continue: 
#compare n1 and n2 to find the largest number
cmp %edx, %ecx
jg n1_maior_n2
    #block else
    cmp %edx, %ebx
    jg firstlarger
        #block else
        sub %edx, %eax
        jmp continue2
    firstlarger:
        sub %ebx, %eax
        jmp continue2

n1_maior_n2:
    cmp %ecx, %ebx
    jg secondlarger
        #block else
        sub %ecx, %eax
        jmp continue2

    secondlarger:
        sub %ebx, %eax #already have the subtraction 
        jmp continue2

continue2:
    pop %edx
    pop %ecx
    pop %ebx
    mov %ebp, %esp
    pop %ebp
    ret

Answer 1

您的代码在做什么:

if (n2 < n1) {
    if (n3 < n2) min = n3;
    else         min = n2;
} else {
    if (n3 < n1) min = n3;
    else         min = n1;
}

if (n2 > n1) {
    if (n3 > n2) return min - n3;
    else         return min - n2;
} else {
    if (n3 > n1) return min - n3;
    else         return min - n1;
}

您可能已经意识到，无论您使用何种语法，这种方法在 x86 ASM 中都不必要地复杂化。 C 代码甚至显示了它!

相反，使用 min 和 max 的默认值并将非默认值与 min 和 进行比较会更有效>max，然后返回减法结果:

int min = n1, max = n3;

if (min > n2) min = n2;
if (min > n3) min = n3;

if (max < n1) max = n1;
if (max < n2) max = n2;

return max - min;

相同数量的必需比较，但请注意 else 分支的消除，这使代码更简单。翻译成一些简单的汇编代码:

    .text
    .globl largestdif
largestdif:
    pushl %ebp
    movl %esp, %ebp

    movl 16(%ebp), %edx  # d = n1, min = n1
    movl 12(%ebp), %ecx  # c = n2
    movl 8(%ebp), %eax   # a = n3, max = n3

# Find min.
    cmpl %ecx, %edx      # if (n1 <= n2) {}
    jle .min_not_n2      # else
    xchgl %ecx, %edx     #   swap(&n1, &n2) // n1 < n2 is true after this
.min_not_n2:
    cmpl %eax, %edx      # if (n1 <= n3) {}
    jle .min_not_n3      # else
    xchgl %eax, %edx     #   swap(&n1, &n3) // n1 < n3 is true after this
.min_not_n3:

# Find max.
    cmpl %ecx, %eax      # if (n3 >= n2) {}
    jge .max_not_n2      # else
    xchgl %ecx, %eax     #   swap(&n2, &n3) // n3 > n2 is true after this
.max_not_n2:

    # n1 > n3 is absolutely false at this point.
    # The swaps above ordered things into the relation
    # n1 <= n2 <= n3 (%edx <= %ecx <= %eax),
    # so n1 <= n3 must be true.

# Return max - min.
    subl %edx, %eax

    movl %ebp, %esp
    popl %ebp
    ret

我已尽力优化它，但我承认几乎可以肯定进一步优化此代码。