Can you please explain this simple increment program to me?

[soche123]

code is

#include <iostream>
#include <conio.h>
using namespace std;

int main()
{
    int count=1;
    while (count++ <=5)
    cout<<count*(count-2)<<" ";
    cout<<endl;

}

WHat I don't get is how count becomes 3,8,15,24...thanks!

[soche123]

another one is

int count=5;
while (count-- > 0)
cout<<count<<" ";
cout<<endl;

Why the output is showing "0" as well when we set the condition of >0?thanks

[tinstaafl]

First off doing increment in a loop is alot easier with a for loop, which is designed for that.

In the first instance, the numbers you are seeing are the result of count*(count-2), not the value of count

In the second instance count is compared to 0 before it is decremented, so when count = 0, it is dercremented before it is compared again.

To re-iterate this kind of confusion is easier to avoid using a for loop whenever you have a closed loop and/or need the incremented value for an expression. to wit:

        for(int count = 2;count <= 6; count++)
        {
             cout<<count*(count-2)<<" ";
        }
            cout<<endl;

produces the same output, but it is a lot easier to see what the values of count will be, namely each integer from 2 to 6 inclusive.

[soche123]

sir thanks for replying..I get about the second one but still confusion about first one..how is

cout<<count*(count-2)<<" "; 

equals 3 after loop runs for the first time?shouldn't it be 2*(2-2)=0?

[nullptr]

What compiler are you using?
The output I get from VS 2012 is: 0 3 8 15 24

[soche123]

i get the same but I don't get logically how it should be 3?

[tinstaafl]

With your original code, the total output is 0 3 8 15 24. The 0 is the result of the first run. Since count starts off equal to 1 then it is incremented so count = 2 and the expression result is 0 (2 * (2 - 2)), the next increment makes count = 3 so the expression will return 3 (3 * (3 - 2)). And so on.

[capslocked]

using G++ (GCC) 4.5.2 the result are:
0 3 8 15 24

I will interpret the code

1) count = 1;
2) loop while count <= 5;
3) count = count + 1;    // the value of count is 2
4) val1 = count - 2;     // the value of val1 is 0
5) val2 = count x val1;  // the value of val2 is 0
6) print val2
7) goto #2

Invadev

[nehal4777]

While loop is working due to when the value of variable less then 5... When variable's value is equal to 5 then the program will automaticaly terminate.

[sanjulovers]

it should be 3 as 3 is the result after second iteration of the loop body. In the second iteration value of count is already 2 when the condition statement is processed. So when entering the loop body, the value of count will get incremented to 3 as count++ is used. And the result is displayed after calculating (count(count - 2)) which is logically equivalent to (3(3-2)) which equals 3.

[RonalBertogi]

He he he. Actually, the OP's confusion is valid. After the state while( count++<=5), count's value will already be 2, thus cout<<count*(count-2)<<" "; will be 2(2-2)==>2(0)==>0. It's like the sum(A1:A5) function in excel returning different result compared to A1+A2+A3+A4+A5.

If you're very much sure there's nothing to check in your codes, then there might be something wrong with your compiler.

[sanjulovers]

Why the problem cannot be with the logic???

[RonalBertogi]

First loop is actually 2x(2-2) which will result to 0. But why is your compiled program returning 3?