古怪：（ - 0.666667 + 0.333333） [英] Weirdness: (-0.666667 + 0.333333)
问题描述
我在一个程序中遇到一些奇怪的东西，当减去2
（double）的时候会导致0 ，但它返回
-1.11022e-16。在我看来，将双x_step decleration
更改为无符号类型可能有所帮助，但编译器在我尝试时会抱怨。
任何想法？
#include< iostream>
使用命名空间std;
int main（int argc，char * argv []）{
double x1 = -1;
double x2 = 1;
double nx = 6;
int pos = 0;
double x = 0.0;
double x_step = 0.0;
x_step =（x2 - x1）/ nx;
x = x1;
for（pos = 0; pos< nx; pos ++）{
cout<< x<< " + \t" << x_step<< " = \t" << （x +
x_step）<< endl;
x = x + x_step;
}
返回0;
}
输出：
-1 + 0.333333 = -0.666667
-0.666667 + 0.333333 = -0.333333
-0.333333 + 0.333333 = -1.11022e-16
-1.11022e-16 + 0.333333 = 0.333333
0.333333 + 0.333333 = 0.666667
0.666667 + 0.333333 = 1
Corne''Cornelius< co **** @ nospam-for-noreason.uy>写在
新闻：P6 ******************** @ is.co.za：
我在程序中遇到一些奇怪的东西，当减去2
（double）'时应该导致0，但它会返回
-1.11022e-16。在我看来，将双x_step decleration更改为unsigned类型可能有所帮助，但编译器在我尝试时会抱怨。
任何想法？
这就是浮点运算的本质。你不应该将
与精确值进行比较，因为舍入错误（这是不可避免的，
虽然通常非常小）。
你应该检查一些小的epsilon范围，而不是比较
的确切数字。
-
:: bartekd [at] o2 [dot] pl
我已将x_step var舍入到十进制后的几个位置
点似乎有帮助，我需要一个接近0的数字然后
溢出的结果。
谢谢！
bartek写道：Corne''Cornelius< co **** @ nospam-for-noreason.uy>在
新闻中写道：P6 ******************** @ is.co.za：
我在程序中遇到一些奇怪的东西，当减去2
（double）'时应该导致0，但它会返回
-1.11022e-16。在我看来，将双x_step decleration更改为unsigned类型可能有所帮助，但编译器在我尝试时会抱怨。
任何想法？
这是浮点运算的本质。因为舍入错误（这是不可避免的，但通常非常小），你不应该与确切的值进行比较。
你应该检查一些小的epsilon范围而不是比较确切的数字。
嗨！
Corne''Cornelius <共**** @ nospam-for-noreason.uy>写道：
我在程序中遇到一些奇怪的东西，当减去2
（double）'时应该得到0，但它返回
-1.11022e-16。在我看来，将双x_step decleration更改为unsigned类型可能有所帮助，但编译器在我尝试时会抱怨。
问题在于计算机上的实数具有有限的准确性。
减去两个（几乎）相等的实数的结果应该不值得信任。这个问题的解决方案通常取决于你想要做什么。
再见，
Chris Dams
Hi,
I''m experiencing some weirdness in a program, when subtracting 2
(double)''s which should result in 0, but instead it returns
-1.11022e-16. It looks to me that changing the double x_step decleration
to unsigned type, might help, but the compiler complains when i try that.
Any ideas ?
#include <iostream>
using namespace std;
int main(int argc, char *argv[]) {
double x1 = -1;
double x2 = 1;
double nx = 6;
int pos = 0;
double x = 0.0;
double x_step = 0.0;
x_step = (x2 - x1) / nx;
x = x1;
for (pos = 0; pos < nx; pos++) {
cout << x << " +\t" << x_step << " =\t " << (x +
x_step) << endl;
x = x + x_step;
}
return 0;
}
Output:
-1 + 0.333333 = -0.666667
-0.666667 + 0.333333 = -0.333333
-0.333333 + 0.333333 = -1.11022e-16
-1.11022e-16 + 0.333333 = 0.333333
0.333333 + 0.333333 = 0.666667
0.666667 + 0.333333 = 1
Corne'' Cornelius <co****@nospam-for-noreason.uy> wrote in
news:P6********************@is.co.za:
Hi,
I''m experiencing some weirdness in a program, when subtracting 2
(double)''s which should result in 0, but instead it returns
-1.11022e-16. It looks to me that changing the double x_step decleration
to unsigned type, might help, but the compiler complains when i try that.
Any ideas ?
That''s the nature of floating point arithmetic. You shouldn''t compare
against exact values, beacause of rounding errors (which are unavoidable,
though usually very small).
You should check against some small epsilon range instead of comparing
exact numbers.
--
:: bartekd [at] o2 [dot] pl
I''ve rounded the x_step var to a couple of places after the decimal
point which seem to help, i need a number closer to 0 then what the
overflow results in.
Thanks !
bartek wrote:Corne'' Cornelius <co****@nospam-for-noreason.uy> wrote in
news:P6********************@is.co.za:Hi,
I''m experiencing some weirdness in a program, when subtracting 2
(double)''s which should result in 0, but instead it returns
-1.11022e-16. It looks to me that changing the double x_step decleration
to unsigned type, might help, but the compiler complains when i try that.
Any ideas ?
That''s the nature of floating point arithmetic. You shouldn''t compare
against exact values, beacause of rounding errors (which are unavoidable,
though usually very small).
You should check against some small epsilon range instead of comparing
exact numbers.
Hi!
Corne'' Cornelius <co****@nospam-for-noreason.uy> writes:
I''m experiencing some weirdness in a program, when subtracting 2
(double)''s which should result in 0, but instead it returns
-1.11022e-16. It looks to me that changing the double x_step decleration
to unsigned type, might help, but the compiler complains when i try that.
The problem is that real numbers on a computer have limited accuracy.
Results of substracting two real numbers that are (almost) equal, should
not be trusted. Solutions to this problem generally depend on what
exactly you want to do.
Bye,
Chris Dams
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