整数分割乘法 [英] Split Multiplication of integers

问题描述

我需要使用两个32位整数作为参数的算法，并返回这些参数的乘法分成另外两个32位整数：32最高位部和32个最低位部

我会尝试：

  uint32_t的P1，P2; //全局变量来保存结果

无效MULT（uint32_t的X，uint32_t的Y）{
    uint64_t中R =（X * Y）;

    P1 = R＆GT;＆GT; 32;
    P2 = R＆安培; 0xFFFFFFFF的;

}
 

虽然它的工作原理 ¹ ，它不能保证的64位整数的机器的存在，既不是利用他们的编译器。

那么，如何解决这个问题的最好方法是什么？

---

注意 ¹ ：事实上，它没有工作，因为我的编译器不支持64位整数

观测：请，避免使用提升

解决方案

只需使用16位数字。

 无效乘法（uint32_t的一个，uint32_t的B，uint32_t的* H，uint32_t的* 1）{
    uint32_t的常量基地= 0x10000的;
    uint32_t的人=一％的基础啊= A /基地，BL = B％的基础上，BH = B /基地;
    * L = A * BL;
    * H =啊* BH;
    uint32_t的RLH = * L /基地+人* BH;
    * H + = RLH /基地;
    RLH = RLH％基地+啊* BL;
    * H + = RLH /基地;
    * L =（RLH％的基础）*基地+ * 1％的基础;
}
 

I need an algorithm that uses two 32-bit integers as parameters, and returns the multiplication of these parameters split into two other 32-bit integers: 32-highest-bits part and 32-lowest-bits part.

I would try:

uint32_t p1, p2; // globals to hold the result

void mult(uint32_t x, uint32_t y){
    uint64_t r = (x * y);

    p1 = r >> 32;
    p2 = r & 0xFFFFFFFF;

}

Although it works¹, it's not guaranteed the existence of 64-bit integers in the machine, neither is the use of them by the compiler.

So, how is the best way to solve it?

---

Note¹: Actually, it didn't work because my compiler does not support 64-bit integers.

Obs: Please, avoid using boost.

解决方案

Just use 16 bits digits.

void multiply(uint32_t a, uint32_t b, uint32_t* h, uint32_t* l) {
    uint32_t const base = 0x10000;
    uint32_t al = a%base, ah = a/base, bl = b%base, bh = b/base;
    *l = al*bl;
    *h = ah*bh;
    uint32_t rlh = *l/base + al*bh;
    *h += rlh/base;
    rlh = rlh%base + ah*bl;
    *h += rlh/base;
    *l = (rlh%base)*base + *l%base;
}

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