iOS开发-常用的数学方法
1、三角函數
double sin(double); 正弦
double cos(double); 余弦
double tan(double); 正切
?2、反三角函數
double asin(double); 反正弦,結果介于[-PI/2, PI/2]
double acos(double); ?反余弦,結果介于[0, PI]
double atan(double); 反正切(主值), 結果介于[-PI/2, PI/2]
double atan2(double, double); 反正切(整圓值), 結果介于[-PI, PI]
3、雙曲三角函數
double sinh (double);?雙曲正弦
double cosh (double);?雙曲余弦
double tanh (double);? 雙曲正切
4、反雙曲三角函數
double asinh(double);?反雙曲正弦
double acosh(double);?反雙曲余弦
double atanh(double);?反雙曲正切
5、指數、對數、開根
double sqrt(double n); 求 n的平方根
double cbrt(double n); 求 n的立方根
double hypot(double, double);?已知直角三角形兩個直角邊長度,求斜邊長度,即兩數平方之和開根
double result = hypot(3, 4) = 5.000000;double pow(double x, double n);??求 x的n次方
double exp(double n);? 求 e=2.71828的n次方
double result = exp(8) = 2980.957987;double exp2(double n);??求 2的n次方?
double result = exp2(8) = 256.000000;double expm1(double n);?求 e=2.71828的n次方 - 1
double result = expm1(8) = 2979.957987;double log(double n); 求 以e為底的n的對數
double result = log(exp(8) = 8.000000;double log10(double n); 求 以10為底的n的對數
double result = log10(100) = 2.000000;double log2(double n);?求 以2為底的n的對數
double result = log2(8) = 3.000000;6、取整和取余
double fabs(double n);?n的絕對值
double copysign(double a,?double b);? 使a的符號和b一樣
extern double ceil(double); 向上取整
double result = ceil(4.1) = 5.0;extern double floor(double);?向下取整
double result = floor(4.9) = 4.0;double rint(double n);? / double nearbyint(double n);?最接近n的整數,一樣近優(yōu)先偶數。
double result = nearbyint(1.5) = 2; double result = nearbyint(2.5) = 2; double result = nearbyint(1.411) = 1;double round(double); 四舍五入
double trunc(double); 去除小數部分
double fmod(double x, double y); x/y取余
double remainder(double, double);?x/y取余,余數可以為負數,取絕對值最小的余數
double result = remainder(0.6, 0.13) = -0.05 0.6 / 0.13 = 4 ... 0.08 0.6 / 0.13 = 5 ... -0.05 (這個絕對值最小)double remquo(double, double, int *); 策略與remainder()一致, 商通過指針回傳
int quotient = 0; double result = remquo(0.6, 0.13, "ient) = -0.05; quotient = 5;double modf(double x, double *y);?將參數x的整數部分通過指針傳給y,返回值為小數部分
double integer = 0; double decimal = modf(3.1415926, &integer); integer = 3.000000; decimal = 0.141593;7、比較
double fdim(double a, double b);? x = a - b <= 0 ? 0 : a - b
double fmax(double, double); 取最大值
double fmin(double, double);?取最小值
double fma(double a, double b, double c);?x = a * b + c
8、標準化浮點數
double frexp (double f, int *p); 標準化浮點數,f = x * 2^p,已知 f 求 x,p ( x介于[0.5, 1] )
double x = frexp(8, &p) = 0.05; p = 4;double ldexp (double x, int p); 與 frexp 相反, 已知 x,p 求 f
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