diff --git a/Python/dtoa.c b/Python/dtoa.c index 9e3c5d82408..9ca2dd95a04 100644 --- a/Python/dtoa.c +++ b/Python/dtoa.c @@ -1672,6 +1672,16 @@ _Py_dg_strtod(const char *s00, char **se) } } else if (e1 < 0) { + /* The input decimal value lies in [10**e1, 10**(e1+16)). + + If e1 <= -512, underflow immediately. + If e1 <= -256, set bc.scale to 2*P. + + So for input value < 1e-256, bc.scale is always set; + for input value >= 1e-240, bc.scale is never set. + For input values in [1e-256, 1e-240), bc.scale may or may + not be set. */ + e1 = -e1; if ((i = e1 & 15)) dval(&rv) /= tens[i]; @@ -1742,7 +1752,34 @@ _Py_dg_strtod(const char *s00, char **se) if (bd0 == NULL) goto failed_malloc; + /* Notation for the comments below. Write: + + - dv for the absolute value of the number represented by the original + decimal input string. + + - if we've truncated dv, write tdv for the truncated value. + Otherwise, set tdv == dv. + + - srv for the quantity rv/2^bc.scale; so srv is the current binary + approximation to tdv (and dv). It should be exactly representable + in an IEEE 754 double. + */ + for(;;) { + + /* This is the main correction loop for _Py_dg_strtod. + + We've got a decimal value tdv, and a floating-point approximation + srv=rv/2^bc.scale to tdv. The aim is to determine whether srv is + close enough (i.e., within 0.5 ulps) to tdv, and to compute a new + approximation if not. + + To determine whether srv is close enough to tdv, compute integers + bd, bb and bs proportional to tdv, srv and 0.5 ulp(srv) + respectively, and then use integer arithmetic to determine whether + |tdv - srv| is less than, equal to, or greater than 0.5 ulp(srv). + */ + bd = Balloc(bd0->k); if (bd == NULL) { Bfree(bd0); @@ -1755,6 +1792,7 @@ _Py_dg_strtod(const char *s00, char **se) Bfree(bd0); goto failed_malloc; } + /* tdv = bd * 10^e; srv = bb * 2^(bbe - scale) */ bs = i2b(1); if (bs == NULL) { Bfree(bb); @@ -1775,6 +1813,17 @@ _Py_dg_strtod(const char *s00, char **se) bb2 += bbe; else bd2 -= bbe; + + /* At this stage e = bd2 - bb2 + bbe = bd5 - bb5, so: + + tdv = bd * 2^(bbe + bd2 - bb2) * 5^(bd5 - bb5) + srv = bb * 2^(bbe - scale). + + Compute j such that + + 0.5 ulp(srv) = 2^(bbe - scale - j) + */ + bs2 = bb2; j = bbe - bc.scale; i = j + bbbits - 1; /* logb(rv) */ @@ -1782,9 +1831,26 @@ _Py_dg_strtod(const char *s00, char **se) j += P - Emin; else j = P + 1 - bbbits; + + /* Now we have: + + M * tdv = bd * 2^(bd2 + scale + j) * 5^bd5 + M * srv = bb * 2^(bb2 + j) * 5^bb5 + M * 0.5 ulp(srv) = 2^bs2 * 5^bb5 + + where M is the constant (currently) represented by 2^(j + scale + + bb2 - bbe) * 5^bb5. + */ + bb2 += j; bd2 += j; bd2 += bc.scale; + + /* After the adjustments above, tdv, srv and 0.5 ulp(srv) are + proportional to: bd * 2^bd2 * 5^bd5, bb * 2^bb2 * 5^bb5, and + bs * 2^bs2 * 5^bb5, respectively. */ + + /* Remove excess powers of 2. i = min(bb2, bd2, bs2). */ i = bb2 < bd2 ? bb2 : bd2; if (i > bs2) i = bs2; @@ -1793,6 +1859,8 @@ _Py_dg_strtod(const char *s00, char **se) bd2 -= i; bs2 -= i; } + + /* Scale bb, bd, bs by the appropriate powers of 2 and 5. */ if (bb5 > 0) { bs = pow5mult(bs, bb5); if (bs == NULL) { @@ -1847,6 +1915,11 @@ _Py_dg_strtod(const char *s00, char **se) goto failed_malloc; } } + + /* Now bd, bb and bs are scaled versions of tdv, srv and 0.5 ulp(srv), + respectively. Compute the difference |tdv - srv|, and compare + with 0.5 ulp(srv). */ + delta = diff(bb, bd); if (delta == NULL) { Bfree(bb);