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RoundScalar() fallback use powf(10.f, -x) instead of 1.0f/powf(10.0f,x)

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ocornut 2015-08-05 17:11:20 -06:00
parent 12d941a42a
commit a54995eace
1 changed files with 1 additions and 1 deletions
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imgui.cpp
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@ -6000,7 +6000,7 @@ static inline float RoundScalar(float value, int decimal_precision)
// So when our value is 1.99999 with a precision of 0.001 we'll end up rounding to 2.0
// FIXME: Investigate better rounding methods
static const float min_steps[10] = { 1.0f, 0.1f, 0.01f, 0.001f, 0.0001f, 0.00001f, 0.000001f, 0.0000001f, 0.00000001f, 0.000000001f };
float min_step = (decimal_precision >= 0 && decimal_precision < 10) ? min_steps[decimal_precision] : (1.0f / powf(10.0f, (float)decimal_precision));
float min_step = (decimal_precision >= 0 && decimal_precision < 10) ? min_steps[decimal_precision] : powf(10.0f, (float)-decimal_precision);
bool negative = value < 0.0f;
value = fabsf(value);
float remainder = fmodf(value, min_step);