浮点数相等 [英] Floating point equality

问题描述

众所周知，比较浮点值时必须小心.通常，我们不使用==，而是使用一些基于epsilon或ULP的相等性测试.

但是，我想知道是否有任何情况，使用==完全可以吗?

看看这个简单的代码片段，可以保证成功的情况是什么?

void fn(float a, float b) {
    float l1 = a/b;
    float l2 = a/b;

    if (l1==l1) { }        // case a)
    if (l1==l2) { }        // case b)
    if (l1==a/b) { }       // case c)
    if (l1==5.0f/3.0f) { } // case d)
}

int main() {
    fn(5.0f, 3.0f);
}

注意:我已经检查了此和当前标准草案中找到的唯一相关声明. :

浮点类型的值表示形式是实现定义的. [注:本文档对浮点运算的准确性没有任何要求；另请参见[support.limits]. —注释]

那么，这是否意味着甚至定义了案例a)"?我的意思是，l1==l1绝对是浮点运算.因此，如果实现是不准确的"，那么l1==l1可能为假吗?

---

我认为这个问题与不是浮点==永远可以吗?.这个问题没有解决我要问的任何情况.相同的主题，不同的问题.我想针对情况a)-d)专门找到答案，为此我无法在重复的问题中找到答案.

解决方案

但是，我想知道是否有任何情况，当使用==完全可以吗?

当然有.一类示例是不涉及计算的用法，例如只能在更改时执行的设置器:

void setRange(float min, float max)
{
    if(min == m_fMin && max == m_fMax)
        return;

    m_fMin = min;
    m_fMax = max;

    // Do something with min and/or max
    emit rangeChanged(min, max);
}

另请参见 浮点==可以吗? 和 是浮点==曾经好吗? .

It is common knowledge that one has to be careful when comparing floating point values. Usually, instead of using ==, we use some epsilon or ULP based equality testing.

However, I wonder, are there any cases, when using == is perfectly fine?

Look at this simple snippet, which cases are guaranteed to succeed?

void fn(float a, float b) {
    float l1 = a/b;
    float l2 = a/b;

    if (l1==l1) { }        // case a)
    if (l1==l2) { }        // case b)
    if (l1==a/b) { }       // case c)
    if (l1==5.0f/3.0f) { } // case d)
}

int main() {
    fn(5.0f, 3.0f);
}

Note: I've checked this and this, but they don't cover (all of) my cases.

Note2: It seems that I have to add some plus information, so answers can be useful in practice: I'd like to know:

• what the C++ standard says
• what happens, if a C++ implementation follows IEEE-754

---

This is the only relevant statement I found in the current draft standard:

The value representation of floating-point types is implementation-defined. [ Note: This document imposes no requirements on the accuracy of floating-point operations; see also [support.limits]. — end note ]

So, does this mean, that even "case a)" is implementation defined? I mean, l1==l1 is definitely a floating-point operation. So, if an implementation is "inaccurate", then could l1==l1 be false?

---

I think this question is not a duplicate of Is floating-point == ever OK?. That question doesn't address any of the cases I'm asking. Same subject, different question. I'd like to have answers specifically to case a)-d), for which I cannot find answers in the duplicated question.

解决方案

However, I wonder, are there any cases, when using == is perfectly fine?

Sure there are. One category of examples are usages that involve no computation, e.g. setters that should only execute on changes:

void setRange(float min, float max)
{
    if(min == m_fMin && max == m_fMax)
        return;

    m_fMin = min;
    m_fMax = max;

    // Do something with min and/or max
    emit rangeChanged(min, max);
}

See also Is floating-point == ever OK? and Is floating-point == ever OK?.

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