实现以下目标最有效的算法是什么:
0010 0000 => 0000 0100
从MSB->LSB转换为LSB->MSB。所有位都必须反转;也就是说,这不是字节交换。
实现以下目标最有效的算法是什么:
0010 0000 => 0000 0100
从MSB->LSB转换为LSB->MSB。所有位都必须反转;也就是说,这不是字节交换。
当前回答
unsigned char ReverseBits(unsigned char data)
{
unsigned char k = 0, rev = 0;
unsigned char n = data;
while(n)
{
k = n & (~(n - 1));
n &= (n - 1);
rev |= (128 / k);
}
return rev;
}
其他回答
我认为这是最简单的逆转比特的方法之一。 如果这个逻辑有什么缺陷,请让我知道。 基本上在这个逻辑中,我们检查位置的位的值。 如果值为1,则在反转位置上设置位。
void bit_reverse(ui32 *data)
{
ui32 temp = 0;
ui32 i, bit_len;
{
for(i = 0, bit_len = 31; i <= bit_len; i++)
{
temp |= (*data & 1 << i)? (1 << bit_len-i) : 0;
}
*data = temp;
}
return;
}
原生ARM指令“rbit”可以用1个cpu周期和1个额外的cpu寄存器来完成,不可能被击败。
似乎许多其他帖子都关心速度(即最好=最快)。 简单性怎么样?考虑:
char ReverseBits(char character) {
char reversed_character = 0;
for (int i = 0; i < 8; i++) {
char ith_bit = (c >> i) & 1;
reversed_character |= (ith_bit << (sizeof(char) - 1 - i));
}
return reversed_character;
}
并希望聪明的编译器将为您优化。
如果你想反转一个更长的位列表(包含sizeof(char) * n位),你可以使用这个函数得到:
void ReverseNumber(char* number, int bit_count_in_number) {
int bytes_occupied = bit_count_in_number / sizeof(char);
// first reverse bytes
for (int i = 0; i <= (bytes_occupied / 2); i++) {
swap(long_number[i], long_number[n - i]);
}
// then reverse bits of each individual byte
for (int i = 0; i < bytes_occupied; i++) {
long_number[i] = ReverseBits(long_number[i]);
}
}
这将把[10000000,10101010]反向转换为[01010101,00000001]。
unsigned char ReverseBits(unsigned char data)
{
unsigned char k = 0, rev = 0;
unsigned char n = data;
while(n)
{
k = n & (~(n - 1));
n &= (n - 1);
rev |= (128 / k);
}
return rev;
}
// Purpose: to reverse bits in an unsigned short integer
// Input: an unsigned short integer whose bits are to be reversed
// Output: an unsigned short integer with the reversed bits of the input one
unsigned short ReverseBits( unsigned short a )
{
// declare and initialize number of bits in the unsigned short integer
const char num_bits = sizeof(a) * CHAR_BIT;
// declare and initialize bitset representation of integer a
bitset<num_bits> bitset_a(a);
// declare and initialize bitset representation of integer b (0000000000000000)
bitset<num_bits> bitset_b(0);
// declare and initialize bitset representation of mask (0000000000000001)
bitset<num_bits> mask(1);
for ( char i = 0; i < num_bits; ++i )
{
bitset_b = (bitset_b << 1) | bitset_a & mask;
bitset_a >>= 1;
}
return (unsigned short) bitset_b.to_ulong();
}
void PrintBits( unsigned short a )
{
// declare and initialize bitset representation of a
bitset<sizeof(a) * CHAR_BIT> bitset(a);
// print out bits
cout << bitset << endl;
}
// Testing the functionality of the code
int main ()
{
unsigned short a = 17, b;
cout << "Original: ";
PrintBits(a);
b = ReverseBits( a );
cout << "Reversed: ";
PrintBits(b);
}
// Output:
Original: 0000000000010001
Reversed: 1000100000000000