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

Question

I know how to obtain the square root of a number using the sqrt function.

How can I obtain the cube root of a number?

Answer 1

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

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

Math for Dummies.

Answer 2

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.

Answer 3

sqrt stands 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.

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

• 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.

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

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 1/3 uses integer division and results in 0.}

Answer 4

in C++11 std::cbrt was introduced as part of math library, you may refer

Answer 5

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

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

Answer 6

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.

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