| // Copyright 2022 Google LLC |
| // |
| // This source code is licensed under the BSD-style license found in the |
| // LICENSE file in the root directory of this source tree. |
| |
| #include <assert.h> |
| #include <stddef.h> |
| |
| #include <arm_neon.h> |
| |
| #include <xnnpack/math-stubs.h> |
| |
| |
| void xnn_math_f16_expm1minus__neonfp16arith_rr1_p3( |
| size_t n, |
| const void* input, |
| void* output) |
| { |
| assert(n % (8 * sizeof(__fp16)) == 0); |
| |
| // The largest x for which expm1f(x) is saturated at -1.0f. |
| const float16x8_t vsat_cutoff = vmovq_n_f16(-0x1.0A4p+3f); |
| // Large number such that ulp(magic bias) == 1 and magic bias === 15 mod 2**9. |
| const float16x8_t vmagic_bias = vmovq_n_f16(0x1.83Cp+10f); |
| const float16x8_t vlog2e = vmovq_n_f16(0x1.714p+0f); |
| const float16x8_t vminus_ln2 = vmovq_n_f16(-0x1.630p-1f); |
| // Coefficient of polynomial approximation |
| // exp(t) - 1 ~ t * (1 + t * (c2 + t * c3)) |
| // on [-log(2)/2, log(2)/2] |
| const float16x8_t vc3 = vmovq_n_f16(0x1.56Cp-3f); |
| const float16x8_t vc2 = vmovq_n_f16(0x1.020p-1f); |
| const float16x8_t vone = vmovq_n_f16(1.0f); |
| |
| const __fp16* i = (const __fp16*) input; |
| __fp16* o = (__fp16*) output; |
| for (; n != 0; n -= 8 * sizeof(__fp16)) { |
| float16x8_t vx = vld1q_f16(i); i += 8; |
| |
| // The function saturates at -1 for large negative inputs: expm1h(x) == -1.0h for x <= sat_cutoff ~= -8.3203125. |
| // To guarantee this behaviour, we clip input at sat_cutoff, and leverage the fact that for our implementation |
| // expm1m(sat_cutoff) == -1.0f. NaN inputs are passed unchanged. |
| vx = vmaxq_f16(vx, vsat_cutoff); |
| |
| // Compute reduced argument n := round(x / log(2)). |
| // We do it by adding a large number (magic bias), which cause rounding of the result to integer, then subtracing |
| // the large number back. The addition is combined with multiplication by log2e into a single FMA instruction. The |
| // trick with adding large number is valid only within certain bounds (|x / log(2)| <= 2**9, i.e. |
| // |x| <= 0x1.630p+8 = 355.0), but that is acceptable, because inputs x are restricted to [-8.3203125, 0]. |
| // Note that addition-subtraction of the large number doesn't cause overflow for inputs in this range. |
| float16x8_t vn = vfmaq_f16(vmagic_bias, vx, vlog2e); |
| |
| // Create a floating-point number s (scale) such that s == 2**n for valid inputs, i.e. |
| // -8.3203125 <= x <= 0.0, and -12 <= n <= 0 accordingly. |
| // For NaN inputs, s would have zero mantissa and can have arbitrary sign and exponent, depending on the input |
| // NaN payload. In these cases, n and t are NaNs with the same payload as input while s is non-NaN, and thus |
| // input payload would be propagated in all computations. |
| const float16x8_t vs = vreinterpretq_f16_s16(vshlq_n_s16(vreinterpretq_s16_f16(vn), 10)); |
| |
| // Subtract the large number back to get final n := round(x / log(2)). |
| vn = vsubq_f16(vn, vmagic_bias); |
| |
| // Compute reduced argument t := x - n * log(2). |
| float16x8_t vt = vfmaq_f16(vx, vn, vminus_ln2); |
| |
| // Compute degree-3 polynomial approximation for exp(t) - 1 on [-log(2)/2, log(2)/2]. |
| // P(t) = t * (1 + t * (c2 + t * c3)) |
| // = t + t * (t * (c2 + t * c3)) = t + t * p |
| float16x8_t vp = vfmaq_f16(vc2, vc3, vt); |
| vp = vmulq_f16(vp, vt); |
| |
| // Reconstruct the exp(x) - 1 value: |
| // exp(x) - 1 = s * (1 + t * (1 + t * (c2 + t * c3))) - 1 |
| // = (s - 1) + s * (t + t * p) |
| // = ((t * s) + (t * s) * p) + (s - 1) |
| vt = vmulq_f16(vt, vs); |
| const float16x8_t vsm1 = vsubq_f16(vs, vone); |
| vp = vfmaq_f16(vt, vp, vt); |
| const float16x8_t vf = vaddq_f16(vp, vsm1); |
| |
| vst1q_f16(o, vf); o += 8; |
| } |
| } |