sqrt, sqrtl and sqrtf in C++

There are various functions available in the C++ Library to calculate the square root of a number. Most prominently, sqrt is used. It takes double as an argument. The <cmath> header defines two more inbuilt functions for calculating the square root of a number (apart from sqrt) which has an argument of type float and long double. Therefore, all the functions used for calculating square root in C++ are:

Function	Datatype
sqrt	double
sqrtf	float
sqrtl	long double

The functions have been discussed in detail below:

A) double sqrt(double arg): It returns the square root of a number to type double. 

Syntax:

double sqrt(double arg)

CPP

// CPP code to illustrate the use of sqrt function
#include <cmath>
#include <iomanip>
#include <iostream>
using namespace std;

// Driver Code
int main()
{
    double val1 = 225.0;
    double val2 = 300.0;

    cout << fixed << setprecision(12) << sqrt(val1) << endl;
    cout << fixed << setprecision(12) << sqrt(val2) << endl;

    return (0);
}

Output

15.000000000000
17.320508075689

Time Complexity: O(√n)
Auxiliary Space: O(1)

Errors and Exceptions Associated with this Function: 

1. It is mandatory to give the argument otherwise, it will give an error no matching function for call to ‘sqrt()’ as shown below, 

CPP

// CPP Program to demonstrate errors in double sqrt()
#include <cmath>
#include <iostream>
using namespace std;

// Driver Code
int main()
{
    double answer;

    answer = sqrt();
    cout << "Square root of " << a << " is " << answer
         << endl;

    return 0;
}

Output

prog.cpp:9:19: error: no matching function for call to ‘sqrt()’
     answer = sqrt();

Time Complexity: O(√n)
Auxiliary Space: O(1)

2. If we pass a negative value in the argument domain error occurs and the output will be the Square root of -a, which is -nan.

CPP

// CPP Program to demonstrate errors in double sqrt()
#include <cmath>
#include <iostream>
using namespace std;

// Driver Code
int main()
{
    double a = -2, answer;

    answer = sqrt(a);
    cout << "Square root of " << a << " is " << answer
         << endl;

    return 0;
}

Output:

Square root of -2 is -nan

Time Complexity: O(√n)
Auxiliary Space: O(1)
B) float sqrtf(float arg): It returns the square root of a number to type float.

Syntax:

float sqrtf(float arg)

CPP

// CPP code to illustrate the use of sqrtf function
#include <cmath>
#include <iomanip>
#include <iostream>

using namespace std;

int main()
{
    float val1 = 225.0;
    float val2 = 300.0;

    cout << fixed << setprecision(12) << sqrtf(val1)
         << endl;
    cout << fixed << setprecision(12) << sqrtf(val2)
         << endl;

    return (0);
}

Output

15.000000000000
17.320508956909

Time Complexity: O(√n)
Auxiliary Space: O(1)
C) long double sqrtl(long double arg): It returns the square root of a number to type long double with more precision. 

Advantage of sqrtl function: When working with integers of the order 10¹⁸, calculating its square root with sqrt function may give an incorrect answer due to precision errors as default functions in programming language works with floats/doubles. But this will always give an accurate answer. 
Syntax:

long double sqrtl(long double arg)

Following is an illustration given below shows the exact difference when working with long integers with sqrt and sqrtl,
1) Using sqrt function:

CPP

// CPP code to illustrate the incorrectness of sqrt
// function
#include <cmath>
#include <iomanip>
#include <iostream>

using namespace std;

int main()
{
    long long int val1 = 1000000000000000000;
    long long int val2 = 999999999999999999;

    cout << fixed << setprecision(12) << sqrt(val1) << endl;
    cout << fixed << setprecision(12) << sqrt(val2) << endl;

    return (0);
}

Output

1000000000.000000000000
1000000000.000000000000

Time Complexity: O(√n)
Auxiliary Space: O(1)

2) Using sqrtl function:

CPP

// CPP code to illustrate the correctness of sqrtl function
#include <cmath>
#include <iomanip>
#include <iostream>

using namespace std;

int main()
{
    long long int val1 = 1000000000000000000;
    long long int val2 = 999999999999999999;

    cout << fixed << setprecision(12) << sqrtl(val1)
         << endl;
    cout << fixed << setprecision(12) << sqrtl(val2)
         << endl;

    return (0);
}

Output

1000000000.000000000000
999999999.999999999476

Time Complexity: O(√n)
Auxiliary Space: O(1)

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.