实现一个32位计算机上的64位运算 [英] implement 64-bit arithmetic on a 32-bit machine

问题描述

以下code计算x和y的乘积，并将结果存储在存储器中。数据
类型ll_t被定义为等同于很长很长。

The following code computes the product of x and y and stores the result in memory. Data type ll_t is defined to be equivalent to long long.

typedef long long ll_t;
void store_prod(ll_t *dest, int x, ll_t y) {
    *dest = x*y;
}

GCC生成以下组装code实现的计算：
DEST在的％ebp + 8，X在的％ebp + 12，Y处的％ebp + 16

gcc generates the following assembly code implementing the computation: dest at %ebp+8, x at %ebp+12, y at %ebp+16

1 movl 16(%ebp), %esi
2 movl 12(%ebp), %eax
3 movl %eax, %edx
4 sarl $31, %edx
5 movl 20(%ebp), %ecx
6 imull %eax, %ecx
7 movl %edx, %ebx
8 imull %esi, %ebx
9 addl %ebx, %ecx
10 mull %esi
11 leal (%ecx,%edx), %edx
12 movl 8(%ebp), %ecx
13 movl %eax, (%ecx)
14 movl %edx, 4(%ecx)

这code使用三个乘法来实现多precision算术
实现的32位机器上的64位算术必需的。描述
算法用于计算产品和注释组装code，显示
它是如何实现你的算法。

This code uses three multiplications to implement the multiprecision arithmetic required to implement 64-bit arithmetic on a 32-bit machine. Describe the algorithm used to compute the product, and annotate the assembly code to show how it realizes your algorithm.

我不明白，在组装$ C $上面C线8和9号线。谁能帮助？的

I don't understand line 8 and line 9 in assembly code above. Can anyone help?

推荐答案

我已经转换它为Intel语法。

I've converted it to intel syntax.

mov esi, y_low
mov eax, x
mov edx, eax
sar edx, 31
mov ecx, y_high

imul ecx, eax ; ecx = y_high *{signed} x

mov ebx, edx

imul ebx, esi ; ebx = sign_extension(x) *{signed} y_low

add ecx, ebx ; ecx = y_high *{signed} x_low + x_high *{signed} y_low

mul esi ; edx:eax = x_low *{unsigned} y_low

lea edx, [ecx + edx] ; edx = high(x_low *{unsigned} y_low + y_high *{signed} x_low + x_high *{signed} y_low)

mov ecx, dest
mov [ecx], eax
mov [ecx + 4], edx

在什么上面的code确实是2的64位有符号整数乘法，保持产品的最低显著64位。

What the above code does is multiplication of 2 64-bit signed integers that keeps the least-significant 64 bits of the product.

在哪里其他64位的被乘数从何而来？这是 X 的符号扩展的从32位到64 特区指令用于复制 X的符号位为 EDX 的所有位。我把这个值只包含了 X的标志 x_high 。 x_low 是 X 实际传递到程序。

Where does the other 64-bit multiplicand come from? It's x sign-extended from 32 bits to 64. The sar instruction is used to replicate x's sign bit into all bits of edx. I call this value consisting only of the x's sign x_high. x_low is the value of x actually passed into the routine.

y_low 和 y_high 是 y的最小，最显著的部分，就像 X的 x_low 和 x_high 是。

从这里，它是pretty容易：

From here it's pretty easy:

产品= 是 * {}签署 X =结果
（ y_high * 2 ³² + y_low ）* {签署}（ x_high * 2 ³² + x_low ）=结果
y_high * {}签署 x_high * 2 ⁶⁴ +结果
y_high * {}签署 x_low * 2 ³² +结果
y_low * {}签署 x_high * 2 ³² +结果
y_low * {}签署 x_low

product = y *{signed} x =
(y_high * 2³² + y_low) *{signed} (x_high * 2³² + x_low) =
y_high *{signed} x_high * 2⁶⁴ +
y_high *{signed} x_low * 2³² +
y_low *{signed} x_high * 2³² +
y_low *{signed} x_low

y_high * {}签署 x_high * 2 ⁶⁴ 不计算，因为它无助于产品的至少显著64位。我们会计算它，如果我们有兴趣在全128位的产品（对于挑剔的全96位产品）。

y_high *{signed} x_high * 2⁶⁴ isn't calculated because it doesn't contribute to the least significant 64 bits of the product. We'd calculate it if we were interested in the full 128-bit product (full 96-bit product for the picky).

y_low * {}签署 x_low 使用无符号乘法计算。这是合法的，这样做是因为2的补符号乘法给出了相同的至少显著位作为无符号乘法。例如：结果
-1 * {}签署-1 = 1结果
0xFFFFFFFFFFFFFFFF * {无符号} 0xFFFFFFFFFFFFFFFF = 0xFFFFFFFFFFFFFFFE0000000000000001（64至少显著位是相当于1）

y_low *{signed} x_low is calculated using unsigned multiplication. It's legal to do so because 2's complement signed multiplication gives the same least significant bits as unsigned multiplication. Example:
-1 *{signed} -1 = 1
0xFFFFFFFFFFFFFFFF *{unsigned} 0xFFFFFFFFFFFFFFFF = 0xFFFFFFFFFFFFFFFE0000000000000001 (64 least significant bits are equivalent to 1)

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