support of __float2bfloat162_rn

[jinz2014]

error: use of undeclared identifier 'float2bfloat162_rn'; did you mean '__float22bfloat162_rn'? return float2bfloat162_rn(a);

Thanks

[cjatin]

Can you share some more information, where are you seeing this error.

[jinz2014]

I hope this link will eventually let you add all missing functions related to conversions among float, half and bfloat16.

https://github.com/ROCm/HIPIFY/blob/amd-staging/docs/tables/CUDA_Device_API_supported_by_HIP.md

[houseroad]

CUDA provided __float2bfloat162_rn, however, hip doesn't have the corresponding definition. So when hipifying the code, it cannot compile.

[yiakwy-xpu-ml-framework-team]

error: use of undeclared identifier 'float2bfloat162_rn'; did you mean '__float22bfloat162_rn'? return float2bfloat162_rn(a);

// Following math functions included in ROCM6.2 SDK : // hmul: bfloat16 -> bfloat16 , hmul2: bfloat16 -> bfloat16, // floats2bfloat162_rn: (float,float) -> __hip_bfloat162, // float22bfloat162_rn: float2 -> hip_bfloat162, // __float2bfloat162_rn : float -> hip_bfloat162, // bfloat1622float2 : hip_bfloat162 -> float2

Here is an simple implementation of __floats2bfloat162_rn:

__inline__ __device__ __nv_bfloat162 __floats2bfloat162_rn(const float a, const float b) {
  __nv_bfloat162 val;
  val = __nv_bfloat162(__float2bfloat16_rn(a), __float2bfloat16_rn(b));
  return val;
}

it means accept two floats, and convert each of them to bf16 and pack it into 32bit number.

The signature '__float2bfloat162_rn' also implemented in hip, it accepts a single float:

__HOST_DEVICE__ inline __hip_bfloat162 __float2bfloat162_rn(const float a) {
  return __hip_bfloat162{__float2bfloat16(a), __float2bfloat16(a)};
}

here is the implemenation:

/**
 * \ingroup HIP_INTRINSIC_BFLOAT16_CONV
 * \brief Converts float to bfloat16
 */
__HOST_DEVICE__ inline __hip_bfloat16 __float2bfloat16(float f) {
  __hip_bfloat16 ret;
  union {
    float fp32;
    unsigned int u32;
  } u = {f};
  if (~u.u32 & 0x7f800000) {
    // When the exponent bits are not all 1s, then the value is zero, normal,
    // or subnormal. We round the bfloat16 mantissa up by adding 0x7FFF, plus
    // 1 if the least significant bit of the bfloat16 mantissa is 1 (odd).
    // This causes the bfloat16's mantissa to be incremented by 1 if the 16
    // least significant bits of the float mantissa are greater than 0x8000,
    // or if they are equal to 0x8000 and the least significant bit of the
    // bfloat16 mantissa is 1 (odd). This causes it to be rounded to even when
    // the lower 16 bits are exactly 0x8000. If the bfloat16 mantissa already
    // has the value 0x7f, then incrementing it causes it to become 0x00 and
    // the exponent is incremented by one, which is the next higher FP value
    // to the unrounded bfloat16 value. When the bfloat16 value is subnormal
    // with an exponent of 0x00 and a mantissa of 0x7F, it may be rounded up
    // to a normal value with an exponent of 0x01 and a mantissa of 0x00.
    // When the bfloat16 value has an exponent of 0xFE and a mantissa of 0x7F,
    // incrementing it causes it to become an exponent of 0xFF and a mantissa
    // of 0x00, which is Inf, the next higher value to the unrounded value.
    u.u32 += 0x7fff + ((u.u32 >> 16) & 1);  // Round to nearest, round to even
  } else if (u.u32 & 0xffff) {
    // When all of the exponent bits are 1, the value is Inf or NaN.
    // Inf is indicated by a zero mantissa. NaN is indicated by any nonzero
    // mantissa bit. Quiet NaN is indicated by the most significant mantissa
    // bit being 1. Signaling NaN is indicated by the most significant
    // mantissa bit being 0 but some other bit(s) being 1. If any of the
    // lower 16 bits of the mantissa are 1, we set the least significant bit
    // of the bfloat16 mantissa, in order to preserve signaling NaN in case
    // the bloat16's mantissa bits are all 0.
    u.u32 |= 0x10000;  // Preserve signaling NaN
  }

  ret.data = (u.u32 >> 16);
  return ret;
}

@houseroad @jinz2014 please updated your SDK to rocm 6.2 using this script

[jinz2014]

Thanks for the answer.