The world is doomed when computers take over!...

Well I did what you told me I used a double int thing. Well ok double var_name. And well I had a user put in 2 numbers (the user was me) and then pick a operation. I only had + sence it was a test. But anyways I added to small numbers 2+2. And guess what it equalled. NOT 4! it was over 2000! I will give the code I used...

Code:

---

#include <iostream.h>
#include <conio.h>
#include <windows.h>
int main()
{
    char operation;
   
    long double number_1;
    long double number_2;
    long double number_3;
    long double number_4;
    long double resault_1;
    long double resault_2;
   
    number_1 = 0;
    number_2 = 0;
    number_3 = 0;
    number_4 = 0;
   
   
    cout<<"Number 1: ";
    cin>>number_1;
   
    cout<<"Operator: ";
    cin>>operation;
   
    cout<<"Number 2: ";
    cin>>number_2;
   
   
    if(operation==1)
        resault_1=number_1+number_2;
     
   
   
    cout<<resault_1;
   
   
   
   
   
    getch();
    return 0;
}

---

I know how I pick the operation is wierd but like I said this is a test. You can try it and see the resaults if you want.


Edit: Why does it do that?

Edit Edit: It also does that if I take away the longs before the doubles.

I did this

Code:

---

#include <iostream>

int main ( void )
{

    long double number_1;
    long double number_2;
    long double result;

    std::cout << "Number 1: ";
    std::cin  >> number_1;

    std::cout << std::endl;
       
    std::cout << "Number 2: ";
    std::cin >> number_2;

    std::cout << std::endl;

    result = number_1 + number_2;

    std::cout << result;

    return 0;

}

---

And I did

Code:

---


Number 1: 2
Number 2: 2
4


---

[edit]
Ah, try operator == '1';

When you input 1 the numeric value is atcually 0x31.
You need the quotes since you declared operator as a char.

This never happened...


But when I put quotes on the 1 it says ISO forbids the delcoration of pointer to integer or something like that.

When it asks you to input the operation... do you type 1?

You know how they say you learn something every day...

Well I learned 2 things...


edit I thought I had to use " instead of '. BUt I found out you have to use '.

Here I fixed it all for you :p

Code:

---

// Standard Header
#include <iostream>
using namespace std;

int main ( void )
{
   
    char operation;
   
    long double number_1;
    long double number_2;

    // Corrected a small typo :)
    long double result_1;
   
    number_1 = 0;
    number_2 = 0;   
   
    cout << "Number 1: ";
    cin  >> number_1;
   
    cout << "Operator: ";
    cin  >> operation;

    cout << "Number 2: ";
    cin  >> number_2;

    cout << endl;
   
   
    if ( operation == '+' )
        result_1 = number_1 + number_2;
     
   
    cout << result_1; 
   
    // Replaced getch() with a standard function
    cin.get();
    return 0;

}

---

There, now it is all standard and pretty :)

[edit]
Oh, your problem was caused because since operation didnt actually equal numeric 1, result was never initialized. So the result was undefined.

[edit2]
Notice I put '+' , so when it asks for operation you literally put +...

can't I use getch()? I like becuase it stands out and I first leanred that and cin.get and getch() do the same thing and no one will know but you guys so why does it matter so much?


But anyways...

Is there a way to prevent the number from going into scientific notaion or what this pic shows. Becuase I want it to show a plain flat out number.

Edit: The number below the 111111111+111111 is what it equals.


Oh and thanks for your help again!

The only way I know of would be to use printf () to format it, because cout will automatically use scientific notation if needed.

BUT, the number would be so large that you wouldnt get an acurate number..

It would be like -53545252 or something...

So im not really sure how to do really large numbers without Scientific Notation.

[edit]
Yeah you can use getch().. just dont tell anybody :p
As long as you understand that its not standard and a really bad habbit. ;)

You can format the output of cout using manipulators in <iomanip>.

And in the code above, I would initialize result_1 to 0 when it is defined because if the operation is wrong you will get whatever the unitialized value was. Since your code allows for this possibilty, then you should intialize the variable. I would also wait to declare/define it until just before it is used - e.g. before the if statement that checks the operation.

nope printf() gives me this error

33 C:\Documents and Settings\Alex\Desktop\Math Helper\main.cpp cannot convert `long double' to `const char*' for argument `1' to `int printf(const


and the peice of code is

printf(resault_1);

If you use printf correctly it won't give you an error.

But use cout anyway as long as you are using it for the rest of your input/output. Just look up io manipulators in your book or tutorial and learn about the different options available to you. Chances are that several of them will help you get the output you want.

I'm googleling it right now...

Well the best I got so far is...

I extended the number by about half more. So it shows about half more then it did before. I am still going for more tho.

Yeah sorry about suggesting printf ();

I didnt even think about iomanip. I dont know why :(

lol oh well. I am still looking into other ways. I already tried a .txt file. didn't work.