原创 在STM32上快速实现8x8点阵的行列变换

【45楼】 minux 啊啊?  给出了更简洁的第二种算法。
示例程序如下：
#include <stdio.h>

void transpose8(unsigned char i[8], unsigned char o[8]) {
        unsigned long x, y, t; 
        x = (i[0] << 24) | (i[1] << 16) | (i[2] << 8) | i[3];
        y = (i[4] << 24) | (i[5] << 16) | (i[6] << 8) | i[7];

        t = (x & 0xf0f0f0f0) | ((y >> 4) & 0x0f0f0f0f);
        y = ((x << 4) & 0xf0f0f0f0) | (y & 0x0f0f0f0f);
        x = t;

        t = (x ^ (x >> 14)) & 0x0000cccc;
        x = x ^ t ^ (t << 14);
        t = (y ^ (y >> 14)) & 0x0000cccc;
        y = y ^ t ^ (t << 14);

        t = (x ^ (x >> 7)) & 0x00aa00aa;
        x = x ^ t ^ (t << 7);
        t = (y ^ (y >> 7)) & 0x00aa00aa;
        y = y ^ t ^ (t << 7);

        o[7] = x >> 24; o[6] = x >> 16; o[5] = x >> 8; o[4] = x;
        o[3] = y >> 24; o[2] = y >> 16; o[1] = y >> 8; o[0] = y;
}

int main(void) {
        int i;
        unsigned char t1[8]={0x24,0x21,0xf0,0x7f,0x80,0x37,0xff,0x1f};
        unsigned char t2[8];

        transpose8(t1, t2);
        for (i = 0; i < 8; i++)
                printf("0x%x ", t2);
        return EOF == putchar('\n');
}