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Reverse bit pattern in C

问题描述:

I am converting a number to binary and have to use putchar to output each number.

The problem is that I am getting the order in reverse.

Is there anyway to reverse a numbers bit pattern before doing my own suff to it?

As in int n has a specific bit pattern - how can I reverse this bit pattern?

网友答案:

There are many ways to do this, some very fast. I had to look it up.

Reverse bits in a byte

b = ((b * 0x0802LU & 0x22110LU) | (b * 0x8020LU & 0x88440LU)) * 0x10101LU >> 16; 

Reverse an N-bit quantity in parallel in 5 * lg(N) operations:

unsigned int v; // 32-bit word to reverse bit order

// swap odd and even bits
v = ((v >> 1) & 0x55555555) | ((v & 0x55555555) << 1);
// swap consecutive pairs
v = ((v >> 2) & 0x33333333) | ((v & 0x33333333) << 2);
// swap nibbles ... 
v = ((v >> 4) & 0x0F0F0F0F) | ((v & 0x0F0F0F0F) << 4);
// swap bytes
v = ((v >> 8) & 0x00FF00FF) | ((v & 0x00FF00FF) << 8);
// swap 2-byte long pairs
v = ( v >> 16             ) | ( v               << 16);

Reverse bits in word by lookup table

static const unsigned char BitReverseTable256[256] = 
{
#   define R2(n)     n,     n + 2*64,     n + 1*64,     n + 3*64
#   define R4(n) R2(n), R2(n + 2*16), R2(n + 1*16), R2(n + 3*16)
#   define R6(n) R4(n), R4(n + 2*4 ), R4(n + 1*4 ), R4(n + 3*4 )
    R6(0), R6(2), R6(1), R6(3)
};

unsigned int v; // reverse 32-bit value, 8 bits at time
unsigned int c; // c will get v reversed

// Option 1:
c = (BitReverseTable256[v & 0xff] << 24) | 
    (BitReverseTable256[(v >> 8) & 0xff] << 16) | 
    (BitReverseTable256[(v >> 16) & 0xff] << 8) |
    (BitReverseTable256[(v >> 24) & 0xff]);

// Option 2:
unsigned char * p = (unsigned char *) &v;
unsigned char * q = (unsigned char *) &c;
q[3] = BitReverseTable256[p[0]]; 
q[2] = BitReverseTable256[p[1]]; 
q[1] = BitReverseTable256[p[2]]; 
q[0] = BitReverseTable256[p[3]];

Please look at http://graphics.stanford.edu/~seander/bithacks.html#ReverseParallel for more information and references.

网友答案:

Pop bits off your input and push them onto your output. Multiplying and dividing by 2 are the push and pop operations. In pseudo-code:

reverse_bits(x) {
    total = 0
    repeat n times {
       total = total * 2
       total += x % 2 // modulo operation
       x = x / 2
    }
    return total
}

See modulo operation on Wikipedia if you haven't seen this operator.

Further points:

  • What would happen if you changed 2 to 4? Or to 10?
  • How does this effect the value of n? What is n?
  • How could you use bitwise operators (<<, >>, &) instead of divide and modulo? Would this make it faster?
  • Could we use a different algorithm to make it faster? Could lookup tables help?
网友答案:

Let me guess: you have a loop that prints the 0th bit (n&1), then shifts the number right. Instead, write a loop that prints the 31st bit (n&0x80000000) and shifts the number left. Before you do that loop, do another loop that shifts the number left until the 31st bit is 1; unless you do that, you'll get leading zeros.

Reversing is possible, too. Somthing like this:

unsigned int n = 12345; //Source
unsigned int m = 0; //Destination
int i;
for(i=0;i<32;i++)
{
    m |= n&1;
    m <<= 1;
    n >>= 1;
}
网友答案:

I know: that is not exactly C, but I think that is an interesting answer:

int reverse(int i) {
  int output;
  __asm__(
     "nextbit:"
        "rcll $1, %%eax;"
        "rcrl $1, %%ebx;"
        "loop nextbit;"
        : "=b" (output)
        : "a" (i), "c" (sizeof(i)*8) );
  return output;
}

The rcl opcode puts the shifted out bit in the carry flag, then rcr recovers that bit to another register in the reverse order.

网友答案:

My guess is that you have a integer and you're attempting to convert it to binary?

And the "answer" is ABCDEFG, but your "answer" is GFEDCBA?

If so, I'd double check the endian of the machine you're doing it on and the machine the "answer" came from.

网友答案:

Here are functions I've used to reverse bits in a byte and reverse bytes in a quad.

inline unsigned char reverse(unsigned char b) {
    return (b&1 << 7)
         | (b&2 << 5)
         | (b&4 << 3)
         | (b&8 << 1)
         | (b&0x10 >> 1)
         | (b&0x20 >> 3)
         | (b&0x40 >> 5)
         | (b&0x80 >> 7);
}

inline unsigned long wreverse(unsigned long w) {
    return ( ( w     &0xFF) << 24)
         | ( ((w>>8) &0xFF) << 16)
         | ( ((w>>16)&0xFF) << 8)
         | ( ((w>>24)&0xFF) );
}
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