How do you write a c plus plus program to calculate the sum of squares using a function?

Answer 1

/**recursive function to find square root of a double to a precision of

maxDepth = 10 decimal places

returns -1.0 when given a negative number*/

#define maxDepth 10

/*firstly find the rational part*/

double getSqrt(double numIn){

/*cant take the square root of a negitive,

and a square root should not be negative

so return -1*/

int candidate = 0;

if (numIn <0)

return -1.0;

/*try every integer until you get one whose square is higher than

the number required, and this clearly one more than the

integer part of your square root*/

do{

if (candidate *candidate *1.0 numIn)

return testVal;

if (testVal * testVal >numIn)

/*most square roots are irrational, and theirfore a maximum number of recursions

must be set, otherwise infinite recursion will occur*/

if (myDepth <maxDepth)

return getIrrational(numIn, testVal-1/10^myDepth, myDepth +1);

else

return testVal-1/10^myDepth;

}

//this can probably be improved on in terms of conciseness, but the logic is the only //way to find a sqrt without going into calculous

Answer 2

Use a template function to calculate the square of any value, then sum the squares. The following example demonstrates how to sum the squares of integers 3 and 5.

#include<iostream>

template <class T>

T sq(T& value)

{

return( value*value );

}

int main()

{

#using namespace std;

int x=3, y=5;

cout<<"The sum of "<<x<<" squared plus "<<y<<" squared is "

<<sq(x)+sq(y)<<endl;

return(0);

}

Answer 3

#include<iostream>

template<typename T>

struct foo

{

T data;

foo (const T num): data (num) {}

T square() const { return data * data; }

T cube() const { return data * square(); }

};

int main()

{

foo<int> a {3};

foo<double> b {3.14};

std::cout << a.data << " squared = " << a.square() << " and cubed = " << a.cube() << std::endl;

std::cout << b.data << " squared = " << b.square() << " and cubed = " << b.cube() << std::endl;

}

Answer 4

#include <iostream>

#include <cmath>

using namespace std;

int main()

{

int x;

cout << "Enter a number: ";

cin >> x;

cout << "The squareroot of " << x << " is " << sqrt(x) << endl;

char wait;

cin >> wait;

return 0;

}

Answer 5

#include <iostream>

using namespace std;

#include <stdlib.h>

int main()

{

double a;

double b;

double c;

cout << "Enter number for a: ";

cin >> a;

cout << "Enter number for b: ";

cin >> b;

cout << "Enter number for c: ";

cin >> c;

if (a 0)

{

cerr << "0 is an invalid input" << endl;

}

else

{

double root_1 = (-b + sqrt(b * b - 4 * a * c)) / (2 * a);

double root_2 = (-b - sqrt(b * b - 4 * a * c)) / (2 * a);

cout << "Root 1: " << root_1 << endl;

cout << "Root 2: " << root_2 << endl << endl;

}

return 0;

}

Answer 6

#include <iostream>

#include <cmath>

using namespace std;

int main()

{

int num = 0;

cout << "Enter a number: ";

cin >> num;

if(num % 2 != 0)

{

cout << "Number is odd and the squareroot is: " << sqrt(num);

}

return 0;

}

Answer 7

#include

#include

#include

int main()

{

using std::string;

using std::cin;

using std::cout;

using std::endl;

using std::sqrt;

using std::getline;

using std::stringstream;

double input;

string str = "";

while (true)

{

cout << "Enter a positive decimal number: ";

getline (cin, str);

stringstream ss (str);

if (ss >> input && input > 0)

break;

cout << "Invalid input, please try again..." << endl;

}

double squared = input * input;

double square_root = sqrt (input);

cout << input << " squared is " << squared << endl;

cout << "The square root of " << input << " is " << square_root << endl;

}

Example output

Enter a positive decimal number: 1.234

1.234 squared is 1.52276

The square root of 1.234 is 1.11086

Answer 8

That's what function sqrt is good for, prototype in math.h.