为什么我从std :: fmod和std :: remainder获得不同的结果 [英] Why am I getting a different result from std::fmod and std::remainder
问题描述
在下面的示例应用中,我用 953
和 0.1
计算浮点余数$ c> std :: fmod
我所期望的是,因为 953.0 / 0.1 == 9530
,
std :: fmod(953,0.1)== 0
0.1 - 为什么会这样?
请注意, $ c> std :: remainder 我得到正确的结果。
这是:
std :: fmod(953,0.1)== 0.1 // unexpected
std :: remainder(953,0.1)== 0 //预期
两个函数之间的区别:
std :: fmod
计算以下内容:正确的值 x - n * y
,其中 n
是 x / y
/ p>
-
std :: remainder
计算如下:
正确的值 x - n * y
,其中 n
是最接近准确值的整数值 x / y
两个函数具有相同的输出。为什么不是这样?
范例应用程式:
#include< iostream>
#include< cmath>
bool is_zero(double in)
{
return std :: fabs(in)< 0.0000001;
}
int main()
{
double numerator = 953;
double denominator = 0.1;
double quotient = numerator / denominator;
double fmod = std :: fmod(molecular,denominator);
double rem = std :: remainder(numerator,denominator);
if(is_zero(fmod))
fmod = 0;
if(is_zero(rem))
rem = 0;
std :: cout<< quotient:<商<< ,fmod:< fmod<< ,rem:< rem< std :: endl;
return 0;
}
输出:
商:9530,fmod:0.1,rem:0
欢迎来到浮点数学。这里是发生了什么:十分之一不能精确地表示在二进制,正如三分之一不能精确地表示十进制。结果,除法产生的结果略低于9530.底操作产生整数9529而不是9530.然后剩下0.1剩余。
In the below example app I calculate the floating point remainder from dividing 953
by 0.1
, using std::fmod
What I was expecting is that since 953.0 / 0.1 == 9530
, that std::fmod(953, 0.1) == 0
I'm getting 0.1
- why is this the case?
Note that with std::remainder
I get the correct result.
That is:
std::fmod (953, 0.1) == 0.1 // unexpected
std::remainder(953, 0.1) == 0 // expected
Difference between the two functions:
According to cppreference.com
std::fmod
calculates the following:
exactly the value x - n*y
, where n
is x/y
with its fractional part truncated
std::remainder
calculates the following:
exactly the value x - n*y
, where n
is the integral value nearest the exact value x/y
Given my inputs I would expect both functions to have the same output. Why is this not the case?
Exemplar app:
#include <iostream>
#include <cmath>
bool is_zero(double in)
{
return std::fabs(in) < 0.0000001;
}
int main()
{
double numerator = 953;
double denominator = 0.1;
double quotient = numerator / denominator;
double fmod = std::fmod (numerator, denominator);
double rem = std::remainder(numerator, denominator);
if (is_zero(fmod))
fmod = 0;
if (is_zero(rem))
rem = 0;
std::cout << "quotient: " << quotient << ", fmod: " << fmod << ", rem: " << rem << std::endl;
return 0;
}
Output:
quotient: 9530, fmod: 0.1, rem: 0
Welcome to floating point math. Here's what happens: One tenth cannot be represented exactly in binary, just as one third cannot be represented exactly in decimal. As a result, the division produces a result slightly below 9530. The floor operation produces the integer 9529 instead of 9530. And then this leaves 0.1 left over.
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