pow() help

**MB1** · 04-18-2005

I keep getting '1' as the answer to this. My instructor says to use
1.0/3.0 instead of 1/3 to get the cuberoot. Then I get an overload error. Help?!

Code:

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

int main()
{
	float x = 20;

	float cuberoot = pow(x, 1/3);
		cout << cuberoot;

	getch();
	return 0;
}

**PJYelton** · 04-18-2005

1/3 is zero in integer math, thus x to the zero power will always be one. I know theres a power function that deals with floats, let me see if I can find it...

**PorkyChop** · 04-18-2005

For my compiler(GCC 3.4.3) mathcalls.h defines pow to take two doubles for arguments. I tested it and the statement:

Code:

pow(20.0, 1.0/3.0);

works just fine.

**major_small** · 04-18-2005

Originally Posted by MB1

I keep getting '1' as the answer to this. My instructor says to use
1.0/3.0 instead of 1/3 to get the cuberoot. Then I get an overload error. Help?!

Code:

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

int main()
{
	float x = 20;

	float cuberoot = pow(x, 1/3);
		cout << cuberoot;

	getch();
	return 0;
}

the overload error is because you're calling pow(float,int) when pow really only accepts (double,double). overloading is when you use the same function name (pow) with different arguments. basically, all you need to do is send two doubles to pow. try this:

Code:

float cuberoot = static_cast<float>(pow(static_cast<double>(x), 1.0/3.0));

the first cast isn't entirely necessary, but it's good practice.

**MB1** · 04-19-2005

Thanks for the help. I got the actual program working.

I was supposed to compute water flow to 1000 feet per second and I couldn't get the cuberoot part towork.

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