C++学习------cmath头文件的源码学习08

函数族定义---四舍五入与余数函数

ceil---返回不小于x的最小整数

double ceil (double x);

代码示例：

printf ( "ceil of 2.3 is %.1f\n", ceil(2.3) );
  printf ( "ceil of 3.8 is %.1f\n", ceil(3.8) );
  printf ( "ceil of -2.3 is %.1f\n", ceil(-2.3) );
  printf ( "ceil of -3.8 is %.1f\n", ceil(-3.8) );
  //测试结果：
  ceil of 2.3 is 3.0 
  ceil of 3.8 is 4.0 
  ceil of -2.3 is -2.0 
  ceil of -3.8 is -3.0

floor---返回不大于x的最大整数

double floor (double x);

代码示例：

printf ( "floor of 2.3 is %.1lf\n", floor (2.3) );
  printf ( "floor of 3.8 is %.1lf\n", floor (3.8) );
  printf ( "floor of -2.3 is %.1lf\n", floor (-2.3) );
  printf ( "floor of -3.8 is %.1lf\n", floor (-3.8) );
  //测试结果：
floor of 2.3 is 2.0
floor of 3.8 is 3.0
floor of -2.3 is -3.0
floor of -3.8 is -4.0

fmod---返回浮点数求余的结果

double fmod (double numer     , double denom);

fmod = numer - tquot * denom，其中tquot = number/denom; 代码示例：

printf ( "fmod of 5.3 / 2 is %f\n", fmod (5.3,2) );
  printf ( "fmod of 18.5 / 4.2 is %f\n", fmod (18.5,4.2) );
  //测试结果：fmod of 5.3 / 2 is 1.300000 
  fmod of 18.5 / 4.2 is 1.700000

注意第二个计算中实际上4.2*4=16.8,这时剩下的18.5-16.8=1.7，不够做除法，当做余数返回

trunc---返回最接近x且绝对值不比x绝对值大的整数

double trunc (     double x);

代码示例：

const char * format = "%.1f \t%.1f \t%.1f \t%.1f \t%.1f\n";
  printf ("value\tround\tfloor\tceil\ttrunc\n");
  printf ("-----\t-----\t-----\t----\t-----\n");
  printf (format, 2.3,round( 2.3),floor( 2.3),ceil( 2.3),trunc( 2.3));
  printf (format, 3.8,round( 3.8),floor( 3.8),ceil( 3.8),trunc( 3.8));
  printf (format, 5.5,round( 5.5),floor( 5.5),ceil( 5.5),trunc( 5.5));
  printf (format,-2.3,round(-2.3),floor(-2.3),ceil(-2.3),trunc(-2.3));
  printf (format,-3.8,round(-3.8),floor(-3.8),ceil(-3.8),trunc(-3.8));
  printf (format,-5.5,round(-5.5),floor(-5.5),ceil(-5.5),trunc(-5.5));
  //测试结果：
value   round   floor   ceil    trunc
-----   -----   -----   ----    -----
2.3     2.0     2.0     3.0     2.0
3.8     4.0     3.0     4.0     3.0
5.5     6.0     5.0     6.0     5.0
-2.3    -2.0    -3.0    -2.0    -2.0
-3.8    -4.0    -4.0    -3.0    -3.0
-5.5    -6.0    -6.0    -5.0    -5.0

可以看到trunc函数实际上是取整数部分，忽略小数部分

round---返回四舍五入的最近整数

double round (double x);

例子同上一个

lround---返回四舍五入的最近整数，以long int返回

long int lround (double x);

llround---返回四舍五入的最近整数，以long long int返回

long long int llround (double x);

rint---使用fegetround函数中的浮点数规则进行舍入

double rint (double x);

该函数可能接收FE_INEXACT的异常 fegetround的使用参考​​### fegetround---读取舍入信息​​ 代码示例：

printf ("rounding using ");
  switch (fegetround()) {
    case FE_DOWNWARD: printf ("downward"); break;
    case FE_TONEAREST: printf ("to-nearest"); break;
    case FE_TOWARDZERO: printf ("toward-zero"); break;
    case FE_UPWARD: printf ("upward"); break;
    default: printf ("unknown");
  }
  printf (" rounding:\n");

  printf ( "rint (2.3) = %.1f\n", rint(2.3) );
  printf ( "rint (3.8) = %.1f\n", rint(3.8) );
  printf ( "rint (-2.3) = %.1f\n", rint(-2.3) );
  printf ( "rint (-3.8) = %.1f\n", rint(-3.8) );
  //测试结果
Rounding using to-nearest rounding:
rint (2.3) = 2.0
rint (3.8) = 4.0
rint (-2.3) = -2.0
rint (-3.8) = -4.0

lrint---使用fegetround函数中的浮点数规则进行舍入，返回long int

long int lrint (double x);

llrint---使用fegetround函数中的浮点数规则进行舍入，返回long long int

long long int llrint (double x);

nearbyint---使用fegetround函数中的浮点数规则进行舍入

double nearbyint (double x);

该函数不会接收FE_INEXACT的异常，与rint有区别

remainder---返回浮点数除法的余数，但是这里的商会先四舍五入到最近的整数，然后再计算余数

double remainder (double numer     , double denom);

与函数fmod相似，但又有不同，来看一看相似的例子：

printf ( "remainder of 5.3 / 2 is %f\n", remainder (5.3,2) );
  printf ( "remainder of 18.5 / 4.2 is %f\n", remainder (18.5,4.2) );
  //测试结果
  remainder of 5.3 / 2 is -0.700000 
  remainder of 18.5 / 4.2 is 1.700000

其中第一个计算5.3/2 = 2.65，所以商舍入到整数3，然后计算余数5.3-2*3 = -0.7

remquo---返回浮点数除法的余数

double remquo (double numer     , double denom     , int* quot);

操作与remainder类似，但是会将商存储在quot中，例子如下：

double numer = 10.3;
  double denom = 4.5;
  int quot;
  double result = remquo (numer,denom,");
  printf ("numerator: %f\n", numer);
  printf ("denominator: %f\n", denom);
  printf ("remainder: %f\n", result);
  printf ("quotient: %d\n", quot);
  //测试结果
numerator: 10.300000
denominator: 4.500000
remainder: 1.300000
quotient: 2

浮点操作函数

copysign---返回一个数，它的绝对值等于x，符号与y相同

double copysign (double x     , double y);

示例：

printf ("copysign ( 10.0,-1.0) = %f\n", copysign( 10.0,-1.0));
  printf ("copysign (-10.0,-1.0) = %f\n", copysign(-10.0,-1.0));
  printf ("copysign (-10.0, 1.0) = %f\n", copysign(-10.0, 1.0));
  //测试结果copysign ( 10.0,-1.0) = -10.0 
  copysign (-10.0,-1.0) = -10.0 
  copysign (-10.0, 1.0) = 10.0

nan---返回一个double类型的NAN数

double nan (const char* tagp);

默认传递""或者"NAN"都可以。

nextafter---返回指定数x在指定数y方向（正数正方向，负数负方向）上所能表示的下一个数

double nextafter (double x     , double y );

示例：

printf ("first representable value greater than zero: %e\n", nextafter(0.0,1.0));
  printf ("first representable value less than zero: %e\n", nextafter(0.0,-1.0));
  //测试结果
  first representable value greater than zero: 4.940656e-324 
  first representable value less than zero: -4.940656e-324

可以看到，大于0的下一个可以表示的数为4.940656e-324，小于0的为-4.940656e-324

nexttoward---与nextafter相同，但y更精确，使用了long double类型

double nexttoward (double x     , long double y);

示例：

printf ("first representable value greater than zero: %e\n", nexttoward(0.0,1.0L));
  printf ("first representable value less than zero: %e\n", nexttoward(0.0,-1.0L));
  //测试结果
first representable value greater than zero: 4.940656e-324
first representable value less than zero: -4.940656e-324