做64位的有效的方式使用旋转32位值 [英] Efficient way of doing 64 bit rotate using 32 bit values

问题描述

我需要旋转使用2 32位寄存器64位值。有没有人碰到这样的一个有效的方式？

I need to rotate a 64 bit value using 2 32 bit registers. Has anyone come across an efficient way of doing this?

推荐答案

好吧，正常的旋转可以这样实现的：

Well, a normal rotate can be implemented like this:

unsigned int rotate(unsigned int bits, unsigned int n) {
    return bits << n | (bits >> (32 - n));
}

所以，这里的在64位实现猜测的32位瓦尔：

So, here's a guess at a 64-bit implementation with 32-bit vars:

void bit_rotate_left_64(unsigned int hi, unsigned int lo, unsigned int n,
                        unsigned int *out_hi, unsigned int *out_lo) {
    unsigned int hi_shift, hi_rotated;
    unsigned int lo_shift, lo_rotated;

    hi_shift = hi << n;
    hi_rotated = hi >> (32 - n);

    lo_shift = lo << n;
    lo_rotated = lo >> (32 - n);

    *out_hi = hi_shift | lo_rotated;
    *out_lo = lo_shift | hi_rotated;
}

基本上，我只是把旋转位从高字和OR-ING与低位字它们，反之亦然。

Basically, I'm just taking the rotated bits from the high word and OR-ing them with the low word, and vice-versa.

下面是一个简单的测试：

Here's a quick test:

int main(int argc, char *argv[]) { 
    /* watch the one move left */
    hi = 0;
    lo = 1;
    for (i = 0; i < 129; i++) {
        bit_rotate_left_64(hi, lo, 1, &hi, &lo);
        printf("Result: %.8x %.8x\n", hi, lo);
    }

    /* same as above, but the 0 moves left */
    hi = -1U;
    lo = 0xFFFFFFFF ^ 1;
    for (i = 0; i < 129; i++) {
        bit_rotate_left_64(hi, lo, 1, &hi, &lo);
        printf("Result: %.8x %.8x\n", hi, lo);
    }
}

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