Mercurial > cpython-lru_cache-weakref
diff Python/dtoa.c @ 58345:5673dc3b6432 legacy-trunk
Issue #7632: Fix a serious wrong output bug for string -> float conversion.
Also remove some now unused variables, and add comments clarifying the
possible outputs of the parsing section of _Py_dg_strtod. Thanks
Eric Smith for reviewing.
| author | Mark Dickinson <dickinsm@gmail.com> |
|---|---|
| date | Sat, 16 Jan 2010 10:44:00 +0000 |
| parents | 92384d7f425a |
| children | 8e179a0fcb5e |
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--- a/Python/dtoa.c Sat Jan 16 05:15:17 2010 +0000 +++ b/Python/dtoa.c Sat Jan 16 10:44:00 2010 +0000 @@ -1340,7 +1340,7 @@ double _Py_dg_strtod(const char *s00, char **se) { - int bb2, bb5, bbe, bd2, bd5, bbbits, bs2, c, dp0, dp1, dplen, e, e1, error; + int bb2, bb5, bbe, bd2, bd5, bbbits, bs2, c, e, e1, error; int esign, i, j, k, nd, nd0, nf, nz, nz0, sign; const char *s, *s0, *s1; double aadj, aadj1; @@ -1349,7 +1349,7 @@ BCinfo bc; Bigint *bb, *bb1, *bd, *bd0, *bs, *delta; - sign = nz0 = nz = dplen = 0; + sign = nz0 = nz = 0; dval(&rv) = 0.; for(s = s00;;s++) switch(*s) { case '-': @@ -1388,11 +1388,8 @@ else if (nd < 16) z = 10*z + c - '0'; nd0 = nd; - dp0 = dp1 = s - s0; if (c == '.') { c = *++s; - dp1 = s - s0; - dplen = 1; if (!nd) { for(; c == '0'; c = *++s) nz++; @@ -1477,6 +1474,32 @@ if (!nd0) nd0 = nd; + + /* Summary of parsing results. The parsing stage gives values + * s0, nd0, nd, e, y and z such that: + * + * - nd >= nd0 >= 1 + * + * - the nd significant digits are in s0[0:nd0] and s0[nd0+1:nd+1] + * (using the usual Python half-open slice notation) + * + * - the absolute value of the number represented by the original input + * string is n * 10**e, where n is the integer represented by the + * concatenation of s0[0:nd0] and s0[nd0+1:nd+1] + * + * - the first significant digit is nonzero + * + * - the last significant digit may or may not be nonzero; (some code + * currently assumes that it's nonzero; this is a bug) + * + * - y contains the value represented by the first min(9, nd) + * significant digits + * + * - if nd > 9, z contains the value represented by significant digits + * with indices in [9, min(16, nd)). So y * 10**(min(16, nd) - 9) + z + * gives the value represented by the first min(16, nd) sig. digits. + */ + k = nd < DBL_DIG + 1 ? nd : DBL_DIG + 1; dval(&rv) = y; if (k > 9) { @@ -1593,15 +1616,18 @@ /* ASSERT(STRTOD_DIGLIM >= 18); 18 == one more than the */ /* minimum number of decimal digits to distinguish double values */ /* in IEEE arithmetic. */ - i = j = 18; - if (i > nd0) - j += dplen; - for(;;) { - if (--j <= dp1 && j >= dp0) - j = dp0 - 1; - if (s0[j] != '0') + + /* Truncate input to 18 significant digits, then discard any trailing + zeros on the result by updating nd, nd0, e and y suitably. (There's + no need to update z; it's not reused beyond this point.) */ + for (i = 18; i > 0; ) { + /* scan back until we hit a nonzero digit. significant digit 'i' + is s0[i] if i < nd0, s0[i+1] if i >= nd0. */ + --i; + if (s0[i < nd0 ? i : i+1] != '0') { + ++i; break; - --i; + } } e += nd - i; nd = i; @@ -1611,8 +1637,8 @@ y = 0; for(i = 0; i < nd0; ++i) y = 10*y + s0[i] - '0'; - for(j = dp1; i < nd; ++i) - y = 10*y + s0[j++] - '0'; + for(; i < nd; ++i) + y = 10*y + s0[i+1] - '0'; } } bd0 = s2b(s0, nd0, nd, y);
