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

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?

c++


Solution 1:[1]

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

Solution 2:[2]

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

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

Math for Dummies.

Solution 3:[3]

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.

Solution 4:[4]

Just to point this out, though we can use both ways but

 long long res = pow(1e9, 1.0/3);
 long long res2 = cbrt(1e9);
 cout<<res<<endl;
 cout<<res2<<endl;

returns

999
1000

So, in order to get the correct results with pow function we need to add an offset of 0.5 with the actual number or use a double data type i.e.

long long res = pow(1e9+0.5, 1.0/3)
double res = pow(1e9, 1.0/3)

more detailed explanation here C++ pow unusual type conversion

Solution 5:[5]

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

The Correct solution would be

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

Solution 6:[6]

The solution for this problem is

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

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

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

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)1/3 in order to get the correct answer

#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

Solution 7:[7]

You can try this C algorithm :

// return a number that, when multiplied by itself twice, makes N. 
unsigned cube_root(unsigned n){
    unsigned a = 0, b;
    for (int c = sizeof(unsigned) * CHAR_BIT / 3 * 3 ; c >= 0; c -= 3) {
        a <<= 1;
        b = 3 * a * (a + 1) + 1;
        if (n >> c >= b)
            n -= b << c, ++a;
    }
    return a;
}

Solution 8:[8]

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++;
}

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1
Solution 2 OrangeDog
Solution 3 Sebastian Redl
Solution 4
Solution 5 Sagar Kumar
Solution 6
Solution 7 Michel
Solution 8 Pranjal

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