Square root slightly off

[zspar]

Hi, I was supposed to write a basic recursive square root function but the values it returns are slightly off. Im not sure whats wrong, and help would be appreciated.

#include<iostream>
#include<cmath>
#include<cfloat>
#include<string>
#include <cstdlib>
using namespace std;

//declare sqroot function calls recursive function newton
double sqroot1(double num );
double newton(double num, double guess);
int main(int argc, char **argv)
{
    if( argc < 2 )
    {
        return -1;
    }
    
    for( int i = 0 ; i < argc ; i ++)
    {
        cout<<"sqroot("<<argv[i+1]<<") is "<< sqroot1(atoi(argv[i+1])) <<endl;
    }
}

double newton(double num, double a)
{
    
    
    if ((abs(a*a - num) <= FLT_EPSILON))
    {
        return a;
    }
    else
    {
        newton(num, (a+num/a)/2 );
    }
    
    
}
double sqroot1(double num)
{
    double sqrt = newton(num,num/2);
    return sqrt;
    
}

[integralfx]

That is to be expected due to if ((abs(a*a - num) <= FLT_EPSILON)). You will only get the precision of a float (typically 7 digits). Use numeric_limits<float>::digits10 to find out how much precision you can get.
http://www.cplusplus.com/reference/limits/numeric_limits/
I think I'm wrong on this.

[mbozzi]

You're missing a return statement on line 34; in the recursive case, control will reach the end of the function without returning a value.

[lastchance]

You need to:
- return a value on line 34 (as mbozzi says above)
- use atof(), not atoi(), for doubles in line 20
- if you use i+1 in line 20, then it needs to be argc-1, not argc in line 18

Code:

#include<iostream>
#include<cmath>
#include<cfloat>
#include<string>
#include <cstdlib>
using namespace std;

//declare sqroot function calls recursive function newton
double sqroot1(double num );
double newton(double num, double guess);
int main(int argc, char **argv)
{
    if( argc < 2 )
    {
        return -1;
    }
    
    for( int i = 0 ; i < argc - 1; i ++)      // <=== argc - 1, not argc
    {
        cout<<"sqroot("<<argv[i+1]<<") is "<< sqroot1(atof(argv[i+1])) <<endl; // <=== atof, not atoi
    }
}

double newton(double num, double a)
{
    
    
    if ((abs(a*a - num) <= FLT_EPSILON))
    {
        return a;
    }
    else
    {
        return newton(num, (a+num/a)/2 );     // <=== return a value (as per mbozzi suggestion)
    }
    
    
}
double sqroot1(double num)
{
    double sqrt = newton(num,num/2);
    return sqrt;
    
}

Running

newton.exe 4 6.25 0.25 25

gives as output

sqroot(4) is 2
sqroot(6.25) is 2.5
sqroot(0.25) is 0.5
sqroot(25) is 5