Add speed_common_functions.cpp

Author: BBuf

speed_common_functions.cpp

@@ -19,4 +19,118 @@ int ClampToInt(int Value, int Min, int Max) {
 	if (Value < Min) return Min;
 	else if (Value > Max) return Max;
 	else return Value;
+}
+
+//函数3: 整数除以255
+//参考: 无
+//简介: 移位
+int Div255(int Value) {
+	return (((Value >> 8) + Value + 1) >> 8);
+}
+
+//函数4: 取绝对值
+//参考: https://oi-wiki.org/math/bit/
+//简介: 比n > 0 ? n : -n 快
+
+int Abs(int n) {
+	return (n ^ (n >> 31)) - (n >> 31);
+	/* n>>31 取得 n 的符号，若 n 为正数，n>>31 等于 0，若 n 为负数，n>>31 等于 - 1
+	若 n 为正数 n^0=0, 数不变，若 n 为负数有 n^-1
+	需要计算 n 和 - 1 的补码，然后进行异或运算，
+	结果 n 变号并且为 n 的绝对值减 1，再减去 - 1 就是绝对值 */
+}
+
+//函数5: 四舍五入
+//参考: 无
+//简介: 无
+double Round(double V)
+{
+	return (V > 0.0) ? floor(V + 0.5) : Round(V - 0.5);
+}
+
+//函数6: 返回-1到1之间的随机数
+//参考: 无
+//简介: 无
+double Rand()
+{
+	return (double)rand() / (RAND_MAX + 1.0);
+}
+
+//函数7: Pow函数的近似计算，针对double类型和float类型
+//参考: http://www.cvchina.info/2010/03/19/log-pow-exp-approximation/
+//参考: http://martin.ankerl.com/2007/10/04/optimized-pow-approximation-for-java-and-c-c/
+//简介: 这个函数只是为了加速的近似计算，有5%-12%不等的误差
+double Pow(double X, double Y)
+{
+	Approximation V = { X };
+	V.X[1] = (int)(Y * (V.X[1] - 1072632447) + 1072632447);
+	V.X[0] = 0;
+	return V.Value;
+}
+
+
+float Pow(float X, float Y)
+{
+	Approximation V = { X };
+	V.X[1] = (int)(Y * (V.X[1] - 1072632447) + 1072632447);
+	V.X[0] = 0;
+	return (float)V.Value;
+}
+
+//函数8: Exp函数的近似计算，针对double类型和float类型
+double Exp(double Y)			//	用联合体的方式的速度要快些
+{
+	Approximation V;
+	V.X[1] = (int)(Y * 1485963 + 1072632447);
+	V.X[0] = 0;
+	return V.Value;
+}
+
+float Exp(float Y)			//	用联合体的方式的速度要快些
+{
+	Approximation V;
+	V.X[1] = (int)(Y * 1485963 + 1072632447);
+	V.X[0] = 0;
+	return (float)V.Value;
+}
+
+// 函数9: Pow函数更准一点的近似计算，但是速度会稍慢
+// http://martin.ankerl.com/2012/01/25/optimized-approximative-pow-in-c-and-cpp/
+// Besides that, I also have now a slower approximation that has much less error
+// when the exponent is larger than 1. It makes use exponentiation by squaring,
+// which is exact for the integer part of the exponent, and uses only the exponent’s fraction for the approximation:
+// should be much more precise with large Y
+
+double PrecisePow(double X, double Y){
+	// calculate approximation with fraction of the exponent
+	int e = (int)Y;
+	Approximation V = { X };
+	V.X[1] = (int)((Y - e) * (V.X[1] - 1072632447) + 1072632447);
+	V.X[0] = 0;
+	// exponentiation by squaring with the exponent's integer part
+	// double r = u.d makes everything much slower, not sure why
+	double r = 1.0;
+	while (e)
+	{
+		if (e & 1)	r *= X;
+		X *= X;
+		e >>= 1;
+	}
+	return r * V.Value;
+}
+
+//函数10: 返回Min到Max之间的随机数
+//参考: 无
+//简介: Min为随机数的最小值，Max为随机数的最大值
+int Random(int Min, int Max){
+	return rand() % (Max + 1 - Min) + Min;
+}
+
+//函数11: 符号函数
+//参考: 无
+//简介: 无
+int sgn(int X){
+	if (X > 0) return 1;
+	if (X < 0) return -1;
+	return 0;
 }