[cff] Implement in-house power of 10
A minimal power only for natural numbers exponents of ten, for portability. Found the idea in Tcl/Tk but wrote it myself after weeks and it turned out being a different implementation, reverse direction, constexpr, etc.
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@ -75,6 +75,37 @@ struct top_dict_values_t : dict_values_t<OPSTR>
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unsigned int FDArrayOffset;
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};
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/* Compile time calculating 10^n for n = 2^i */
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constexpr double
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pow10_of_2i (unsigned int n)
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{
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return n == 1 ? 10. : pow10_of_2i (n >> 1) * pow10_of_2i (n >> 1);
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}
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static const double powers_of_10[] =
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{
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pow10_of_2i (0x100),
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pow10_of_2i (0x80),
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pow10_of_2i (0x40),
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pow10_of_2i (0x20),
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pow10_of_2i (0x10),
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pow10_of_2i (0x8),
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pow10_of_2i (0x4),
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pow10_of_2i (0x2),
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pow10_of_2i (0x1),
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};
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/* Works for x < 512 */
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inline double
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_hb_pow10 (unsigned int x)
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{
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unsigned int mask = 0x100; /* Should be same with the first element */
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unsigned long result = 1;
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const double *power = powers_of_10;
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for (; mask; ++power, mask >>= 1) if (mask & x) result *= *power;
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return result;
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}
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struct dict_opset_t : opset_t<number_t>
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{
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static void process_op (op_code_t op, interp_env_t<number_t>& env)
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@ -137,7 +168,7 @@ struct dict_opset_t : opset_t<number_t>
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value = (double) (neg ? -int_part : int_part);
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if (frac_count > 0)
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{
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double frac = (frac_part / pow (10.0, (double) frac_count));
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double frac = frac_part / _hb_pow10 (frac_count);
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if (neg) frac = -frac;
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value += frac;
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}
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@ -153,9 +184,9 @@ struct dict_opset_t : opset_t<number_t>
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if (exp_part != 0)
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{
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if (exp_neg)
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value /= pow (10.0, (double) exp_part);
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value /= _hb_pow10 (exp_part);
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else
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value *= pow (10.0, (double) exp_part);
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value *= _hb_pow10 (exp_part);
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}
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return value;
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