A function that takes an int but returns a double

[bluechimera]

hi,
I am having a problem getting the integer I pass into a function to return as double. it is still truncating the value on the return. any suggestions?


watered down version of my code

 
int choice = 0;
{
cout << "Enter a number: ";
cin >> choice;
 
cout << "The new number is: " << endl;
 
return 0;
}
 
 
double oneDivide(int n)
{
return 1/n;
}
 
 

I keep getting zero as the result (truncating the decimal and value after decimal. I cannot use any local variables
Any suggestions ?
Thank you!

[sicarie]

Aside of the fact that you never call oneDivide, I'd say that your compiler is probably doing an optimization on that - it takes an int and divides it by another int, and makes the result an int. Try using 1.0/n

[bluechimera]

Aside of the fact that you never call oneDivide, I'd say that your compiler is probably doing an optimization on that - it takes an int and divides it by another int, and makes the result an int. Try using 1.0/n

ah! ya I didnt really post the proper code. I wrote it quickly before leaving work

I tried this solution and it worked wonderfully! thank you !

[sicarie]

No problem, feel free to post again if you run into something else.

[weaknessforcats]

Better yet, is to not mix integers and floating point.

Either you will lose accuracy or suffer truncation and loss of data. Like when you try to convert a double to an int and the compiler won't do it.

Stick with all integer or all double.

[hsn]

you have to assign n to a double value
double d=n;
thne do what ever you want to do.