在 C++ 中将 FP32 转换为 Bfloat16

Question

如何在 C++ 中将浮点数（1 位符号、8 位表达式、23 位尾数）转换为 Bfloat16（1 位符号、8 位表达式、7 位尾数）？

Answer 1

memcpy wouldn't compile for me in the little endian case for some reason. This is my solution. I have it as a struct here so that I can easily access the data and run through different ranges of values to confirm that it works properly.

struct bfloat16{
   unsigned short int data;
   public:
   bfloat16(){
      data = 0;
   }
   //cast to float
   operator float(){
      unsigned int proc = data<<16;
      return *reinterpret_cast<float*>(&proc);
   }
   //cast to bfloat16
   bfloat16& operator =(float float_val){
      data = (*reinterpret_cast<unsigned int *>(&float_val))>>16;
      return *this;
   }
};

//an example that enumerates all the possible values between 1.0f and 300.0f
using namespace std;

int main(){
   bfloat16 x;
   for(x = 1.0f; x < 300.0f; x.data++){
      cout<<x.data<<" "<<x<<endl;
   }
   
   return 0;
}

Answer 2

As demonstrated in the answer by Botje it is sufficient to copy the upper half of the float value since the bit patterns are the same. The way it is done in that answer violates the rules about strict aliasing in C++. The way around that is to use memcpy to copy the bits.

static inline tensorflow::bfloat16 FloatToBFloat16(float float_val)
{
    tensorflow::bfloat16 retval;
#if __BYTE_ORDER__ == __ORDER_BIG_ENDIAN__
    memcpy(&retval, &float_val, sizeof retval);
#else
    memcpy(&retval, reinterpret_cast<char *>(&float_val) + sizeof float_val - sizeof retval, sizeof retval);
#endif
    return retval;
}

If it's necessary to round the result rather than truncating it, you can multiply by a magic value to push some of those lower bits into the upper bits.

float_val *= 1.001957f;

Answer 3

From the Tensorflow implementation :

static inline tensorflow::bfloat16 FloatToBFloat16(float float_val) {
#if __BYTE_ORDER__ == __ORDER_BIG_ENDIAN__
    return *reinterpret_cast<tensorflow::bfloat16*>(
        reinterpret_cast<uint16_t*>(&float_val));
#else
    return *reinterpret_cast<tensorflow::bfloat16*>(
        &(reinterpret_cast<uint16_t*>(&float_val)[1]));
#endif
}