将浮点数与没有epsilon的0.0进行比较是否可以？ [英] Is it ok to compare floating points to 0.0 without epsilon?

问题描述

我知道，要比较两个浮点值，需要使用一些epsilon精度，因为它们并不精确。但是，我想知道是否存在不需要我的epsilon的极端情况。

I am aware, that to compare two floating point values one needs to use some epsilon precision, as they are not exact. However, I wonder if there are edge cases, where I don't need that epsilon.

我特别想知道做某事是否总是安全的像这样：

In particular, I would like to know if it is always safe to do something like this:

double foo(double x){
    if (x < 0.0) return 0.0;
    else return somethingelse(x); // somethingelse(x) != 0.0
}

int main(){
   int x = -3.0;
   if (foo(x) == 0.0) { 
     std::cout << "^- is this comparison ok?" << std::endl; 
   }
}

我知道有更好的方式写 foo （例如，另外返回一个标志），但是我想知道将 0.0 分配给浮点数是否一般变量，然后将其与 0.0 进行比较。

I know that there are better ways to write foo (e.g. returning a flag in addition), but I wonder if in general is it ok to assign 0.0 to a floating point variable and later compare it to 0.0.

或更笼统，以下比较是否总是正确？

Or more general, does the following comparison yield true always?

double x = 3.3;
double y = 3.3;
if (x == y) { std::cout << "is an epsilon required here?" << std::endl; }

当我尝试使用它时，它似乎起作用了，但可能不应该依赖那。

When I tried it, it seems to work, but it might be that one should not rely on that.

推荐答案

是的，如果返回 0.0 ，则可以将其与 0.0 ; 0可以精确表示为浮点值。如果返回 3.3 ，则必须更加小心，因为 3.3 不能精确表示，因此从例如，使用double浮点将产生不同的值。

Yes, if you return 0.0 you can compare it to 0.0; 0 is representable exactly as a floating-point value. If you return 3.3 you have to be a much more careful, since 3.3 is not exactly representable, so a conversion from double to float, for example, will produce a different value.

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