从C中的字节数组中提取最后N位

英勇的努力!感谢您展示您的尝试。

我在你的版本中添加了一些评论:

unsigned char array[] = { 0xF0, 0xCC, 0xAA, 0xF0}; 
int i = 0; 
int j;
int totalBits = sizeof(array) * 8;  // currently meaning 32
int startBit = totalBits - 5;  // meaning 27 where '5' is the magic number wanted

for (j = startBit; j < totalBits; j++) 
{
            i = i << 1;
            i = i | (array[j] & 1); // Oops! there is no array[27]! Top element is array[3]!
}    

以下是一个似乎有效的版本的粗略草案:

int main(void) {
    unsigned char arr[] = { 0xF0, 0xCC, 0xAA, 0xF0 }; // Your array (add more! Try it out!)

    union { // providing for up to N = 64bits (on my system)
        unsigned char c[8];
        unsigned long l;
    } foo;

    foo.l = 0; // initialise

    size_t sz = sizeof arr / sizeof arr[0]; // source byte count
    size_t n = 0; // destination byte count

    // wastefully copy as many bytes as are available until 'foo' is a full as possible
    // Notice the 'endian-ness' of index n going from 0 to 8.
    // Other hardware could change this to count downward, instead.
    for( size_t i = sz; i && n < sizeof foo; ) {
        foo.c[ n++ ] = arr[ --i ]; // grab one byte
        printf( "%x\n", foo.l ); // debugging
    }

    int N = 5;
    foo.l &= (1<<N)-1; // Mask off the low order N bits from that long

    printf( "%x\n", foo.l );

    return 0;
}

希望这能有所帮助。

逾期警告:的最大值 N ,在此代码中,为 63 …应该加上检查 N 不超过蓄能器的宽度 foo ,如果 N 正是这个数字,绕过移位和掩码操作。。。

附言:当玩比特操作时,使用它通常更安全 unsigned 数据类型。一些C实现显然反对符号位被不适当地篡改。