1、三角函數(shù)

double sin(double); 正弦

double cos(double); 余弦

double tan(double); 正切

?2、反三角函數(shù)

double asin(double); 反正弦，結(jié)果介于[-PI/2, PI/2]

double acos(double); ?反余弦,結(jié)果介于[0, PI]

double atan(double); 反正切(主值), 結(jié)果介于[-PI/2, PI/2]

double atan2(double, double); 反正切(整圓值), 結(jié)果介于[-PI, PI]

3、雙曲三角函數(shù)

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(shù)

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(shù)、對數(shù)、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(shù)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(shù)

double result = log(exp(8) = 8.000000;

double log10(double n); 求 以10為底的n的對數(shù)

double result = log10(100) = 2.000000;

double log2(double n);?求 以2為底的n的對數(shù)

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(shù)，一樣近優(yōu)先偶數(shù)。

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(shù)部分

double fmod(double x, double y); x/y取余

double remainder(double, double);?x/y取余，余數(shù)可以為負(fù)數(shù)，取絕對值最小的余數(shù)

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, &quotient) = -0.05; quotient = 5;

double modf(double x, double *y);?將參數(shù)x的整數(shù)部分通過指針傳給y，返回值為小數(shù)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(biāo)準(zhǔn)化浮點數(shù)

double frexp (double f, int *p); 標(biāo)準(zhǔn)化浮點數(shù)，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

總結(jié)

以上是生活随笔為你收集整理的iOS开发-常用的数学方法的全部內(nèi)容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網(wǎng)站內(nèi)容還不錯，歡迎將生活随笔推薦給好友。