如何在 C++ 中获得立方根？

How can I obtain the cube root in C++?

提问by Hava Darabi

I know how to obtain the square rootof a number using the sqrtfunction.

我知道如何使用该函数获取数字的平方根sqrt。

How can I obtain the cube rootof a number?

如何获得一个数的立方根？

回答by Lightness Races in Orbit

sqrtstands for "square root", and "square root" means raising to the power of 1/2. There is no such thing as "square root with root 2", or "square root with root 3". For other roots, you change the first word; in your case, you are seeking how to perform cube rooting.

sqrt代表“平方根”，“平方根”的意思是提高到 的幂1/2。没有“根为 2 的平方根”或“根为 3 的平方根”这样的东西。对于其他词根，您更改第一个单词；在您的情况下，您正在寻找如何执行立方根。

Before C++11, there is no specific function for this, but you can go back to first principles:

在 C++11 之前，没有具体的函数，但你可以回到最初的原则：

• Square root: std::pow(n, 1/2.)(or std::sqrt(n))
• Cube root: std::pow(n, 1/3.)(or std::cbrt(n)since C++11)
• Fourth root: std::pow(n, 1/4.)
• etc.

• 平方根：（std::pow(n, 1/2.)或std::sqrt(n)）
• 立方根：（std::pow(n, 1/3.)或std::cbrt(n)自 C++11 起）
• 第四根： std::pow(n, 1/4.)
• 等等。

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If you're expecting to pass negative values for n, avoid the std::powsolution — it doesn't support negative inputs with fractional exponents, and this is why std::cbrtwas added:

如果您希望为 传递负值n，请避免使用该std::pow解决方案 -它不支持带有小数指数的负输入，这就是std::cbrt添加的原因：

std::cout << std::pow(-8, 1/3.) << '\n';  // Output: -nan
std::cout << std::cbrt(-8)      << '\n';  // Output: -2

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^{N.B. That .is really important, because otherwise 1/3uses integer division and results in 0.}

^{注意这.真的很重要，因为否则1/3使用整数除法并导致0.}

回答by triclosan

in C++11 std::cbrtwas introduced as part of mathlibrary, you may refer

在 C++11std::cbrt中作为数学库的一部分被引入，你可以参考

回答by OrangeDog

include <cmath>
std::pow(n, 1./3.)

Also, in C++11 there is cbrtin the same header.

此外，在 C++11 中有cbrt相同的头文件。

Math for Dummies.

傻瓜数学。

回答by Sebastian Redl

The nth root of x is equal to x^(1/n), so use std::pow. But I don't see what this has to with operator overloading.

x 的第 n 个根等于 x^(1/n)，因此使用std::pow. 但我不明白这与运算符重载有什么关系。

回答by SpongeBobFan

1. sqrt()is not an operator.
2. You cannot overload sqrt().

1. sqrt()不是运营商。
2. 你不能超载sqrt()。

Please explain why do you need to overload sqrt()so we can help you to do what you want.

请解释为什么您需要超载，sqrt()以便我们可以帮助您做您想做的事。

回答by Pranjal

I would discourage any of the above methods as they didn't work for me. I did pow(64, 1/3.) along with pow(64, 1./3.) but the answer I got was 3
Here's my logic.

我不鼓励上述任何方法，因为它们对我不起作用。我做了 pow(64, 1/3.) 和 pow(64, 1./3.) 但我得到的答案是 3
这是我的逻辑。

ans = pow(n, 1/3.);
if (pow(ans, 3) != n){
   ans++;
}

回答by Sagar Kumar

Actually the round must go for the above solutions to work.

实际上，必须进行回合才能使上述解决方案起作用。

The Correct solution would be

正确的解决方案是

ans = round(pow(n, 1./3.));

ans = round(pow(n, 1./3.));

回答by Akash Vichare

The solution for this problem is

这个问题的解决方案是

cube_root = pow(n,(float)1/3);

and you should #include <math.h>library file

你应该#include <math.h>库文件

Older standards of C/C++ don't support cbrt() function.

较旧的 C/C++ 标准不支持 cbrt() 函数。

When we write code like cube_root = pow(n,1/3);the compiler thinks 1/3 = 0(division problem in C/C++), so you need to do typecasting using (float)1/3in order to get the correct answer

当我们像cube_root = pow(n,1/3);编译器所想的那样编写代码时1/3 = 0（C/C++ 中的除法问题），因此您需要进行类型转换 using(float)1/3才能得到正确答案

#include<iostream.h>
#include<conio.h>
#include<math.h>
using namespace std;

int main(){
float n = 64 , cube_root ;
clrscr();
cube_root = pow(n , (float)1/3);
cout<<"cube root = "<<cube_root<<endl;
getch();
return 0;
}

cube root = 4

立方根 = 4