Add Float16Array #491
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@@ -28,6 +28,7 @@ | |
| #include <stdlib.h> | ||
| #include <string.h> | ||
| #include <inttypes.h> | ||
| #include <math.h> | ||
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| #if defined(_WIN32) | ||
| #include <windows.h> | ||
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@@ -356,6 +357,81 @@ static inline void inplace_bswap32(uint8_t *tab) { | |
| put_u32(tab, bswap32(get_u32(tab))); | ||
| } | ||
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| static inline double fromfp16(uint16_t v) { | ||
| double d, s; | ||
| int e; | ||
| if ((v & 0x7C00) == 0x7C00) { | ||
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There was a problem hiding this comment. Personal preference probably, but I'd define this like FP6_INF or something, for readability... There was a problem hiding this comment. I thought about that but... for doubles there's |
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| d = (v & 0x3FF) ? NAN : INFINITY; | ||
| } else { | ||
| d = (v & 0x3FF) / 1024.; | ||
| e = (v & 0x7C00) >> 10; | ||
| if (e == 0) { | ||
| e = -14; | ||
| } else { | ||
| d += 1; | ||
| e -= 15; | ||
| } | ||
| d = scalbn(d, e); | ||
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There was a problem hiding this comment. If there's a way to implement fromfp16() and tofp16() without using functions from math.h, I'd love to hear it! My initial approach used plain bit shifting but I couldn't get rounding to work correctly in all cases. Ex. test262 expects For anyone interested, here's my first attempt: static inline double fromfp16(uint16_t v) {
union {
uint16_t v;
struct {
uint16_t f:10;
uint16_t e:5;
uint16_t s:1;
} x;
} h;
union {
double v;
struct {
uint64_t f:52;
uint64_t e:11;
uint64_t s:1;
} x;
} q;
h.v = v;
if (h.x.e == 31) { // +/-Infinity, NaN
q.x.f = h.x.f > 0;
q.x.e = 2047;
} else if (h.x.e == 0 && h.x.f == 0) { // TODO subnormal if h.x.f > 0
q.x.f = 0;
q.x.e = 0;
} else {
q.x.f = (uint64_t)h.x.f << 42;
q.x.e = (h.x.e - 15) + 1023;
}
q.x.s = h.x.s;
return q.v;
}
static inline uint16_t tofp16(double d) {
union {
uint16_t v;
struct {
uint16_t f:10;
uint16_t e:5;
uint16_t s:1;
} x;
} h;
union {
double v;
struct {
uint64_t f:52;
uint64_t e:11;
uint64_t s:1;
} x;
} q;
q.v = d;
if (q.x.e == 2047)
return 0x7C00 | (q.x.s << 15) | (q.x.f > 0); // +/-Infinity, NaN
if (d < -65504 || d > 65504)
return 0x7C00 | (q.x.s << 15); // out of range, return Infinity
if (q.x.e == 0 && q.x.f == 0) {
return 0;
}
h.x.f = q.x.f >> 42;
h.x.e = (q.x.e - 1023) + 15;
h.x.s = q.x.s;
return h.v;
}There was a problem hiding this comment. If we really want to avoid the dependency on math.h for this very function perhaps we can borrow it from musl? https://git.musl-libc.org/cgit/musl/tree/src/math/scalbn.c There was a problem hiding this comment. Never mind, I see we use more math functions below. |
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| } | ||
| s = (v & 0x8000) ? -1.0 : 1.0; | ||
| return d * s; | ||
| } | ||
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| static inline uint16_t tofp16(double d) { | ||
| uint16_t f, s; | ||
| double t; | ||
| int e; | ||
| s = 0; | ||
| if (copysign(1, d) < 0) { // preserve sign when |d| is negative zero | ||
| d = -d; | ||
| s = 0x8000; | ||
| } | ||
| if (isinf(d)) | ||
| return s | 0x7C00; | ||
| if (isnan(d)) | ||
| return s | 0x7C01; | ||
| if (d == 0) | ||
| return s | 0; | ||
| d = 2 * frexp(d, &e); | ||
| e--; | ||
| if (e > 15) | ||
| return s | 0x7C00; // out of range, return +/-infinity | ||
| if (e < -25) { | ||
| d = 0; | ||
| e = 0; | ||
| } else if (e < -14) { | ||
| d = scalbn(d, e + 14); | ||
| e = 0; | ||
| } else { | ||
| d -= 1; | ||
| e += 15; | ||
| } | ||
| d *= 1024.; | ||
| f = (uint16_t)d; | ||
| t = d - f; | ||
| if (t < 0.5) | ||
| goto done; | ||
| if (t == 0.5) | ||
| if ((f & 1) == 0) | ||
| goto done; | ||
| // adjust for rounding | ||
| if (++f == 1024) { | ||
| f = 0; | ||
| if (++e == 31) | ||
| return s | 0x7C00; // out of range, return +/-infinity | ||
| } | ||
| done: | ||
| return s | (e << 10) | f; | ||
| } | ||
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| static inline int isfp16nan(uint16_t v) { | ||
| return (v & 0x7FFF) > 0x7C00; | ||
| } | ||
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| static inline int isfp16zero(uint16_t v) { | ||
| return (v & 0x7FFF) == 0; | ||
| } | ||
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| /* XXX: should take an extra argument to pass slack information to the caller */ | ||
| typedef void *DynBufReallocFunc(void *opaque, void *ptr, size_t size); | ||
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Where does one find the rules you followed here? (trying to inform myself and do a good review 😅 )
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I mostly derived it from first principles; they're identical to floats and doubles, except with smaller exponents and mantissas (5 and 10 bits resp. with an exponent bias of 15. Compare doubles: 11 and 52 bits, bias of 1023.)
To get the floating point value, you compute
pow(2, exponent-15) * (1 + mantissa/1024.0). Compare doubles:pow(2, exponent-1023) * (1 + mantissa/pow(2, 52))To go from double to half-float or vice versa... my intuition was that
exponent - 1023 + 15andmantissa >> (52 - 10)would be Good Enough(TM) but alas, it wasn't, so now I'm mostly computing it in the double-precision domain.The Wikipedia page is okay. (edit: what I call 'mantissa', wikipedia calls 'fraction')
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Oh, and if you're worried about correctness: test262 has pretty good coverage so it should be okay. Everything passes.
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Thank you!