Ignoring character data types.

[vani917]

Hello everyone. I was, for a while, solving this problem our teacher had given us. But unfortunately, I keep failing. Our teacher had said that our program is not supposed to accept character data types whenever entered. But it keeps accepting. I would really be grateful if you could help me. Its a code that accepts building types(blg) then accepts kwh present(kwhpres) and previous(kwhprev) and compute for the respective payment(am). Here is my code:

#include<iostream.h>
void main()
{
double kwhpres, kwhprev, am, k, blg;


cout<<" Building Code"<<endl;
cout<<"----------------"<<endl;
cout<<"(1)-Residential"<<endl;
cout<<"(2)-Commercial"<<endl;
cout<<"(3)-Industrial"<<endl;
cout<<"Enter your choice: ";
cin>>blg;

if (blg==1)
{{
cout<<"Enter present reading ";
cin>>kwhpres;
cout<<"Enter previous reading ";
cin>>kwhprev;
k=kwhpres-kwhprev;
}
{if (k<=50&&k>=0)
{
am=27;
cout<<"Amount due: "<<am<<endl;
}

else if (k>50&&k<=100)
{
am=27+((k-50)*0.75);
cout<<"Amount due: "<<am<<endl;
}
else if (k>100&&k<=150)
{
am=27+((k-100)*3.50);
cout<<"Amount due: "<<am<<endl;
}
else if (k>150)
{
am=27+((k-150)*5.50);
cout<<"Amount due: "<<am<<endl;
}
else
{
cout<<"Invalid"<<endl;
}}}

else if (blg==2)
{ {
cout<<"Enter present reading ";
cin>>kwhpres;
cout<<"Enter previous reading ";
cin>>kwhprev;
k=kwhpres-kwhprev;
}
{if (k<=50&&k>=0)
{
am=35;
cout<<"Amount due: "<<am<<endl;
}

else if (k>50&&k<=100)
{
am=35+((k-50)*0.75);
cout<<"Amount due: "<<am<<endl;
}
else if (k>100&&k<=150)
{
am=35+((k-100)*3.50);
cout<<"Amount due: "<<am<<endl;
}
else if (k>150)
{
am=35+((k-150)*5.50);
cout<<"Amount due: "<<am<<endl;
}
else
{
cout<<"Invalid"<<endl;
}}}
else if (blg==3)
{ {
cout<<"Enter present reading ";
cin>>kwhpres;
cout<<"Enter previous reading ";
cin>>kwhprev;
k=kwhpres-kwhprev;
}
{if (k<=50&&k>=0)
{
am=52;
cout<<"Amount due: "<<am<<endl;
}
else if (k>50&&k<=100)
{
am=52+((k-50)*0.75);
cout<<"Amount due: "<<am<<endl;
}
else if (k>100&&k<=150)
{
am=52+((k-100)*3.50);
cout<<"Amount due: "<<am<<endl;
}
else if (k>150)
{
am=52+((k-150)*5.50);
cout<<"Amount due: "<<am<<endl;
}
else
{
cout<<"Invalid"<<endl;
}
}}}

[MrHutch]

You'll need some sort of while loop in there to validate the input and stop the program from progressing until a number has been entered.

Something like:

int my_int;

cout << "Please enter an integer number: ";

while(!(cin >> my_int))
{
    cout << "Invalid input, try again: ";
    cin.clear();
    cin.ignore(1000, '\n');
}

You'll need to adapt it for your code. If you need extra conditions (say, you don't want a negative number to be entered), just add them as or conditions for that while statement.

[o3hC5Di1]

Hi there,

Would you please be so kind to wrap that code in code-tags ([code ]//code here[ /code]) and apply aome indentation for better readability?

As to your issue, there are some ideas on how to achieve this here:
http://stackoverflow.com/questions/5655142/how-to-check-if-input-is-numeric-in-c

All the best,
NwN

[vani917]

Thank you very much. But where specifically in my program should I put the loop part. Im sorry, Im just a bigginer in this proglang. Thanks.

[o3hC5Di1]

Hi there,

Using iHutch's example:

double kwhpres, kwhprev, am, k, blg;


cout<<" Building Code"<<endl;
cout<<"----------------"<<endl;
cout<<"(1)-Residential"<<endl;
cout<<"(2)-Commercial"<<endl;
cout<<"(3)-Industrial"<<endl;
cout<<"Enter your choice: ";

while(!(cin >> big))
{
    cout << "Invalid input, try again: ";
    cin.clear();
    cin.ignore(1000, '\n');
}

All the best,
NwN

[vani917]

So Its before the if else statements. Thank you very much.

[o3hC5Di1]

Indeed, because you need to have the big value first before you can evaluate it in the if/else statements.

All the best,
NwN