可以精确表示为浮点数/双精度的整数范围 [英] Range of integers that can be expressed precisely as floats / doubles

问题描述

可以表示为double（相当于float？）的（连续）整数的确切范围是什么我所问的原因是因为我很好奇这样的问题，如果发生精确度的错误。

What is the exact range of (contiguous) integers that can be expressed as a double (resp. float?) The reason I ask is because I am curious for questions such as this one when a loss of accuracy will occur.

那是

1. 什么是最小正整数 m ，以便 m + 1 不能精确地表示为double（分别为float）？


2. 最大负整数 -n 是什么，使 -n-1 不能精确地表示为double（分别为float）？ （可能与上述相同）。

1. What is the least positive integer m such that m+1 cannot be precisely expressed as a double (resp. float)?
2. What is the greatest negative integer -n such that -n-1 cannot be precisely expressed as a double (resp. float)? (May be the same as the above).

这意味着 -n 和 m 具有精确的浮点表示。我基本上正在寻找两个浮动和双打的范围 [ - n，m] 。

This means that every integer between -n and m has an exact floating-point representation. I'm basically looking for the range [-n, m] for both floats and doubles.

让我们限制标准IEEE 754 32位和64位浮点表示的范围。我知道float有24位精度，double有53位（两者都有一个隐藏的引导位），但是由于浮点表示的复杂性，我正在寻找一个权威的答案。请不要挥手！

Let's limit the scope to the standard IEEE 754 32-bit and 64-bit floating point representations. I know that the float has 24 bits of precision and the double has 53 bits (both with a hidden leading bit), but due to the intricacies of the floating point representation I'm looking for an authoritative answer for this. Please don't wave your hands!

（理想的答案将证明从 0 到 m 是可表达的，而 m + 1 不是。）

(Ideal answer would prove that all the integers from 0 to m are expressible, and that m+1 is not.)

推荐答案

由于您在询问IEEE浮点类型，因此语言无关紧要。

Since you're asking about IEEE floating-point types, the language does not matter.

#include <iostream>
using namespace std;

int main(){

    float f0 = 16777215.; // 2^24 - 1
    float f1 = 16777216.; // 2^24
    float f2 = 16777217.; // 2^24 + 1

    cout << (f0 == f1) << endl;
    cout << (f1 == f2) << endl;

    double d0 = 9007199254740991.; // 2^53 - 1
    double d1 = 9007199254740992.; // 2^53
    double d2 = 9007199254740993.; // 2^53 + 1

    cout << (d0 == d1) << endl;
    cout << (d1 == d2) << endl;
}

输出：

0
1
0
1

所以float的限制是2 ^ 24。而双倍的限制是2 ^ 53。否定的是相同的，因为唯一的区别是符号位。

So the limit for float is 2^24. And the limit for double is 2^53. Negatives are the same since the only difference is the sign bit.

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