readfloat.c: correctly round fraction via division by exact 10^n

The previous code computed `frac_part * pow10_negative[n]`, where
pow10_negative[n] is already a rounded approximation of 10^-n (since
10^-n is not exactly representable in binary). The multiplication then
adds another rounding step, leaving up to ~1 ulp of error.

Dividing `frac_part` by `pow10_positive[n]` is exact for n <= 22 (the
range where 10^n fits exactly in a double), so the division is the
only rounding and the result is correctly rounded. For example,
"0.3".to_f now matches the 0.3 literal's bit pattern (and libc strtod).

src/readfloat.c

@@ -168,10 +168,11 @@ mrb_read_float(const char *str, char **endp, double *fp)
     res = (double)int_part;
   }
   else {
-    // Fast path: combine integer and fractional parts
+    // Divide by the exact 10^n (exact for n <= 22) rather than multiplying
+    // by the inexact 10^-n, so the fraction is correctly rounded.
     res = (double)int_part;
     if (frac_digits > 0) {
-      res += (double)frac_part * mrb_pow10(-frac_digits);
+      res += (double)frac_part / mrb_pow10(frac_digits);
     }
   }