Program based on function overloading

Question

I made a program based on function overloading consisting of 2 functions int cube (int) and float cube(float) . My main function reads values of int x and float y respectively.Now when I run the program I put a float value instead the integer va;ue And when I put 2.5(decimal value) instead of integer value in variable "x" , The compiler does not ask me the value of y and automatically 2 for x (int) and 0.5 for y (float) and returns the cube of 0.5. Why this is happenning. Why 0.5 is automatically stored in y instead of asking input?

My program is like that -

#include<iostream>
using namespace std;
int main()
{
    int x;
    float y;
    int cube(int);
    float cube(float);
    cout<<"Enter a number to find its cube"<<endl;
    cin>>x;
    cout<<"The cube of the number "<<x<<" is "<<cube(x)<<endl;

    cout<<"Enter a number to find its cube"<<endl;
    cin>>y;
    cout<<"The cube of the number "<<y<<" is "<<cube(y)<<endl;
    return 0;
}
int cube (int num)
{
    return num*num*num;
}

float cube (float num)
{
    return num*num*num;
}

Output is -

Enter a number to find its cube
2.5
The cube of number 2 is 8
Enter the number to find its cube
The cube of number 0.5 is 0.125

Can anyone help me about that Thanks

The question title does not really give any clue about the problem. This problem must be reasonably common among beginners, but I couldn't find a dupe with a quick search. — Walter
– Walter, Commented May 27, 2017 at 13:45

Some programmer dude · Accepted Answer · 2017-05-27 13:37:15Z

1

You try to read an int value, but give a floating point value as input. That means the program will read the integer part, stop reading as soon as it sees something that doesn't match the pattern of an integer value (the '.' in your case) and leave it in the input buffer for the next input operation.

If you want to read whole lines and discard unparsed input, then either use std::istream::ignore after each input. Or read a whole line into a string using std::getline and use an std::istringstream to "parse" the input.

answered May 27, 2017 at 13:37

Some programmer dude

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Comments

Walter · Accepted Answer · 2017-05-27 13:37:43Z

0

This is quite simple:

cout<<"Enter a number to find its cube"<<endl;

you enter 2.5

cin>>x;

this reads 2 (sets x=2) and stops, because the .5 cannot be part of an int.

cout<<"The cube of the number "<<x<<" is "<<cube(x)<<endl;

cout<<"Enter a number to find its cube"<<endl;
cin>>y;

the .5 is still in the input stream, so is read to set y=0.5. There is no need to enter another number, and so the program does not stop to wait for input.

answered May 27, 2017 at 13:37

Walter

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