我有这样一种情况,需要将一个动态大小的字节数组放大3倍。
例如:
我已经使用这里的算法生成了一个查找表。 https://dev59.com/7l_Va4cB1Zd3GeqPWLq6#9044057
例如:
10101010 11001100
to
11100011 10001110 00111000 11111100 00001111 11000000
我已经使用这里的算法生成了一个查找表。 https://dev59.com/7l_Va4cB1Zd3GeqPWLq6#9044057
static const uint32_t bitExpandTable[256] = {
00000000, 0x000007, 0x000038, 0x00003f, 0x0001c0, 0x0001c7, 0x0001f8, 0x0001ff,
0x000e00, 0x000e07, 0x000e38, 0x000e3f, 0x000fc0, 0x000fc7, 0x000ff8, 0x000fff,
0x007000, 0x007007, 0x007038, 0x00703f, 0x0071c0, 0x0071c7, 0x0071f8, 0x0071ff,
0x007e00, 0x007e07, 0x007e38, 0x007e3f, 0x007fc0, 0x007fc7, 0x007ff8, 0x007fff,
0x038000, 0x038007, 0x038038, 0x03803f, 0x0381c0, 0x0381c7, 0x0381f8, 0x0381ff,
0x038e00, 0x038e07, 0x038e38, 0x038e3f, 0x038fc0, 0x038fc7, 0x038ff8, 0x038fff,
0x03f000, 0x03f007, 0x03f038, 0x03f03f, 0x03f1c0, 0x03f1c7, 0x03f1f8, 0x03f1ff,
0x03fe00, 0x03fe07, 0x03fe38, 0x03fe3f, 0x03ffc0, 0x03ffc7, 0x03fff8, 0x03ffff,
0x1c0000, 0x1c0007, 0x1c0038, 0x1c003f, 0x1c01c0, 0x1c01c7, 0x1c01f8, 0x1c01ff,
0x1c0e00, 0x1c0e07, 0x1c0e38, 0x1c0e3f, 0x1c0fc0, 0x1c0fc7, 0x1c0ff8, 0x1c0fff,
0x1c7000, 0x1c7007, 0x1c7038, 0x1c703f, 0x1c71c0, 0x1c71c7, 0x1c71f8, 0x1c71ff,
0x1c7e00, 0x1c7e07, 0x1c7e38, 0x1c7e3f, 0x1c7fc0, 0x1c7fc7, 0x1c7ff8, 0x1c7fff,
0x1f8000, 0x1f8007, 0x1f8038, 0x1f803f, 0x1f81c0, 0x1f81c7, 0x1f81f8, 0x1f81ff,
0x1f8e00, 0x1f8e07, 0x1f8e38, 0x1f8e3f, 0x1f8fc0, 0x1f8fc7, 0x1f8ff8, 0x1f8fff,
0x1ff000, 0x1ff007, 0x1ff038, 0x1ff03f, 0x1ff1c0, 0x1ff1c7, 0x1ff1f8, 0x1ff1ff,
0x1ffe00, 0x1ffe07, 0x1ffe38, 0x1ffe3f, 0x1fffc0, 0x1fffc7, 0x1ffff8, 0x1fffff,
0xe00000, 0xe00007, 0xe00038, 0xe0003f, 0xe001c0, 0xe001c7, 0xe001f8, 0xe001ff,
0xe00e00, 0xe00e07, 0xe00e38, 0xe00e3f, 0xe00fc0, 0xe00fc7, 0xe00ff8, 0xe00fff,
0xe07000, 0xe07007, 0xe07038, 0xe0703f, 0xe071c0, 0xe071c7, 0xe071f8, 0xe071ff,
0xe07e00, 0xe07e07, 0xe07e38, 0xe07e3f, 0xe07fc0, 0xe07fc7, 0xe07ff8, 0xe07fff,
0xe38000, 0xe38007, 0xe38038, 0xe3803f, 0xe381c0, 0xe381c7, 0xe381f8, 0xe381ff,
0xe38e00, 0xe38e07, 0xe38e38, 0xe38e3f, 0xe38fc0, 0xe38fc7, 0xe38ff8, 0xe38fff,
0xe3f000, 0xe3f007, 0xe3f038, 0xe3f03f, 0xe3f1c0, 0xe3f1c7, 0xe3f1f8, 0xe3f1ff,
0xe3fe00, 0xe3fe07, 0xe3fe38, 0xe3fe3f, 0xe3ffc0, 0xe3ffc7, 0xe3fff8, 0xe3ffff,
0xfc0000, 0xfc0007, 0xfc0038, 0xfc003f, 0xfc01c0, 0xfc01c7, 0xfc01f8, 0xfc01ff,
0xfc0e00, 0xfc0e07, 0xfc0e38, 0xfc0e3f, 0xfc0fc0, 0xfc0fc7, 0xfc0ff8, 0xfc0fff,
0xfc7000, 0xfc7007, 0xfc7038, 0xfc703f, 0xfc71c0, 0xfc71c7, 0xfc71f8, 0xfc71ff,
0xfc7e00, 0xfc7e07, 0xfc7e38, 0xfc7e3f, 0xfc7fc0, 0xfc7fc7, 0xfc7ff8, 0xfc7fff,
0xff8000, 0xff8007, 0xff8038, 0xff803f, 0xff81c0, 0xff81c7, 0xff81f8, 0xff81ff,
0xff8e00, 0xff8e07, 0xff8e38, 0xff8e3f, 0xff8fc0, 0xff8fc7, 0xff8ff8, 0xff8fff,
0xfff000, 0xfff007, 0xfff038, 0xfff03f, 0xfff1c0, 0xfff1c7, 0xfff1f8, 0xfff1ff,
0xfffe00, 0xfffe07, 0xfffe38, 0xfffe3f, 0xffffc0, 0xffffc7, 0xfffff8, 0xffffff,
};
我试过使用查找表(looping through the byte array using the LUT)来将前3个字节(memcpy the first 3 bytes)复制到新数组中。然而,我的输出似乎从未正确。
有没有人对如何高效实现这个有建议?这将在嵌入式ARM处理器上运行。
编辑
LUT测试
uint8_t msg[] = { 0xaa, 0x02, 0x43, 0x5a, 0x8d, 0x06, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0xd0, 0x84, 0xc6, 0x2d, 0x00, 0xb9 };
uint8_t expanded_msg[63] = { 0 };
uint8_t tmp_val = 0;
uint32_t lut_val;
for (int i = 0; i < 21; i++)
{
tmp_val = *(uint8_t*)(&msg + i);
lut_val = bitExpandTable[tmp_val];
memcpy(&expanded_msg[(i * 3)], &lut_val, 3);
}
print_binary(&msg, 21);
print_binary(&expanded_msg, sizeof(expanded_msg));
输出
[ 10101010 00000010 01000011 01011010 10001101 00000110 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 11010000 10000100 11000110 00101101 00000000 10111001 ]
[ 00111000 10001110 11100011 00000000 00000000 00000000 11000111 10001111 00000011 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000111 01110000 00011100 00000111 11110000 00000011 11111111 11111111 00011111 11000111 11110001 11100011 00000000 00000000 00000000 11111111 11111111 11111111 11000000 10000001 11100011 00000000 00000000 00000000 00000111 01110000 00011100 00111000 10000000 00011111 11111111 11111111 00011111 11000111 11110001 11100011 00000000 00000000 00000000 11111111 11111111 11111111 ]
pdep+(shl+or)*2
可以轻松完成此操作。 - EOF