如何使用log10正确计算整数长度? [英] How does using log10 correctly calculate the length of a integer?

问题描述

int length = (int) floor( log10 (float) number ) + 1;

我的问题本质上是一个数学问题:为什么要取一个数字的log10()，对该数字取底，加1，然后将其强制转换为一个int来正确计算数字的长度?

My question is essentially a math question: WHY does taking the log10() of a number, flooring that number, adding 1, and then casting it into an int correctly calculate the length of number?

我真的很想知道更深的数学解释！

I really want to know the deep mathematical explanation please!

推荐答案

对于具有n位数字的整数number，其值在10^(n - 1)(包括)和10^n之间，因此log10(number)在n - 1(包括)和n之间.然后，函数floor减少小数部分，结果保留为n - 1.最后，在其上添加1即可得到位数.

For an integer number that has n digits, it's value is between 10^(n - 1)(included) and 10^n, and so log10(number) is between n - 1(included) and n. Then the function floor cuts down the fractional part, leaves the result as n - 1. Finally, adding 1 to it gives the number of digits.

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