Changing one bit in double

问题:

I am trying to change one bit in a double so that for example:

Double x: -1.500912597 which is:

Binary: 10111111 11111000 00000011 10111100 11101101 01100100 01001111 10010011

Change one bit in the underlying binary code (for example, bit 16) so that:

Binary: 10111111 11111001 00000011 10111100 11101101 01100100 01001111 10010011

Double x: -1.563412596999999903

Is there some C++ code I can use for this?

回答1:

#include <ieee754.h>

ieee754_double d = {-1.500912597};
d.ieee.mantissa1 |= 1u << 16; // set bit 16 of mantissa
double x = d.d;

Header ieee754.h on my system comes from glibc-headers package.

回答2:

The only portable way is to use memcpy (yes, I know what you're thinking, and no it's not inefficient).

Note that this solution does not take into account byte-ordering. You'd need to cater for that too to be strictly portable.

#include <cstdlib>
#include <cstring>
#include <utility>
#include <iostream>

// only compile this function if Integer is the same size as a double
template<class Integer, std::enable_if_t<sizeof(Integer) == sizeof(double)>* = nullptr>
double or_bits(double input, Integer bits)
{
  Integer copy;

  // convert the double to a bit representation in the integer 
  std::memcpy(&copy, &input, sizeof(input));
  // perform the bitwise op
  copy |= bits;
  // convert the bits back to a double
  std::memcpy(&input, &copy, sizeof(copy));
  // return the double
  return input;
}

int main()
{
  double d = 1.0;
  d = or_bits(d, 0x10ul);
  std::cout << d << std::endl;
}

assembly output on gcc5.3:

double or_bits<unsigned long, (void*)0>(double, unsigned long):
    movq    %xmm0, %rax
    orq     %rdi, %rax
    movq    %rax, -8(%rsp)
    movsd   -8(%rsp), %xmm0
    ret

回答3:

There are endless possibilities for bit manipulation. So you should practice; but I'll give you some examples.

Below code will "set" only bit 3 of x (where bit 0 is the least significant bit):

#define SET_BIT3 (0x08)
x |= SET_BIT3;

Below code will "reset" only bit 3:

#define RESET_BIT3 (0xFFFFFFFFFFFFFFF7)
x &= RESET_BIT3;

Another thing you could do is make a union like this:

typedef union
{
    struct
    {
        unsigned BIT0:1;
        unsigned BIT1:1;
        unsigned BIT2:1;
        unsigned BIT3:1;
        unsigned BIT4:1;
        unsigned BIT5:1;
        unsigned BIT6:1;
        unsigned BIT7:1;
        unsigned BIT8:1;
        unsigned BIT9:1;
        unsigned BIT10:1;
        unsigned BIT11:1;
        unsigned BIT12:1;
        unsigned BIT13:1;
        unsigned BIT14:1;
        unsigned BIT15:1;
    };
    struct
    {
        unsigned BYTE0:8;
        unsigned BYTE1:8;
    };
    struct
    {
        unsigned ALL:16;
    };
}two_bytes;

What you can do with it is this:

two_bytes var;

var.ALL = 0; // clear all bits
var.BYTE1 = 0xFF; // make all bits of most significant byte 1
var.BIT7 = 1; // set only bit 7 

Or you can turn into bitwise operators again:

#define SET_BIT3 (0x08)
var.ALL |= SET_BIT3;