Conversion from int to double

[edman]

I am trying to convert the ouput of this recursive function to double, but no matter what I try the output remains integer. Could someone please assist.

int main()
{
	int x,y;
	cout << "Please enter the low value of the range to add: " << endl;
	cin >> x;
	cout << "Please enter the high value of the range to add: " << endl;
	cin >> y;
	cout << "The sum of the range is: " << static_cast<double>(addRange(x,y)) << endl;
	cin.get();//To keep console window open
	cin.get();//To keep console window open
	return 0;
}

double addRange(int x, int y)
{
	if (x > y)	
		return (0);
	else
		return static_cast<double>(addRange(static_cast<double>(x) + 1,y)) + x;
}

[Nick Evan]

Why all the horrible casting?
Why not just change the function to take in doubles?

double addRange(double x, double y)
{

[edman]

This is an assignment question and we were spcifically asked to return a double from this recursive function that passes two int parameters. It is senseless, I know :-)

[Narue]

>no matter what I try the output remains integer
If you perform an operation on two integers that wouldn't have any precision even if converted to a double, why would you expect anything but a zero precision? No amount of casting will create something that isn't there.

[ivailosp]

#include <iostream>
using namespace std;
double addRange(int x, int y);
int main() {
	int x, y;
	cout << "Please enter the low value of the range to add: " << endl;
	cin >> x;
	cout << "Please enter the high value of the range to add: " << endl;
	cin >> y;
	cout << "The sum of the range is: " << addRange(x, y) << endl;
	cin.get();//To keep console window open
	cin.get();//To keep console window open
	return 0;
}

double addRange(int x, int y) {
	if (x > y)
		return (0);
	else
		return (addRange(x + 1, y)) + x;
}

lol, works fine...

[Narue]

>lol, works fine...
Yes, but I imagine the OP wants the precision to be printed as well. Input of 1 and 5 should print 15.000000 rather than 15, for example. To do that has nothing to do with the type and everything to do with how it's printed:

cout<<"The sum of the range is: "<< fixed << addRange(x, y) <<endl;

[edman]

Thank you for the help. I agree there is no point to the conversion exercise except to teach me something else I didn't know.

[iamthwee]

This sounds pretty silly to me too.