获取int的位数的方法？

本文翻译自：Way to get number of digits in an int?

获取int的长度是否比这种方法更简洁？

int length = String.valueOf(1000).length();

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#1楼

参考：https://stackoom.com/question/5TwF/获取int的位数的方法

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#2楼

How about plain old Mathematics? 普通的老数学怎么样？ Divide by 10 until you reach 0. 除以10直到达到0。

public static int getSize(long number) {
        int count = 0;
        while (number > 0) {
            count += 1;
            number = (number / 10);
        }
        return count;
    }

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#3楼

The logarithm is your friend: 对数是你的朋友：

int n = 1000;
int length = (int)(Math.log10(n)+1);

NB: only valid for n > 0. 注意：仅对n> 0有效。

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#4楼

Since the number of digits in base 10 of an integer is just 1 + truncate(log10(number)) , you can do: 由于整数的基数10中的位数仅为1 + truncate（log10（数字）） ，因此您可以：

public class Test {

    public static void main(String[] args) {

        final int number = 1234;
        final int digits = 1 + (int)Math.floor(Math.log10(number));

        System.out.println(digits);
    }
}

Edited because my last edit fixed the code example, but not the description. 编辑，因为我的上一次编辑修复了代码示例，但没有修改说明。

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#5楼

Your String-based solution is perfectly OK, there is nothing "un-neat" about it. 你的基于String的解决方案完全没问题，没有什么“不整齐”的。 You have to realize that mathematically, numbers don't have a length, nor do they have digits. 你必须在数学上认识到，数字没有长度，也没有数字。 Length and digits are both properties of a physical representation of a number in a specific base, ie a String. 长度和数字都是特定基数中数字的物理表示的属性，即字符串。

A logarithm-based solution does (some of) the same things the String-based one does internally, and probably does so (insignificantly) faster because it only produces the length and ignores the digits. 基于对数的解决方案（基于字符串的）在内部执行（某些）相同的操作，并且可能更快（无关紧要）更快，因为它只生成长度并忽略数字。 But I wouldn't actually consider it clearer in intent - and that's the most important factor. 但我实际上并不认为它的意图更清晰 - 这是最重要的因素。

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#6楼

Can I try? 我可以试试吗？ ;) ;）

based on Dirk's solution 基于Dirk的解决方案

final int digits = number==0?1:(1 + (int)Math.floor(Math.log10(Math.abs(number))));