Will loop ever terminate???

[kphilux]

Hi i am trying to discuss wether this loop will ever terminate any help would be much appreciated!

signed int i=1
while (i>0)
i=i*1000;

[Deathcry]

no since i will always be greater than 0

[WingedPanther]

Yes, under most languages. When you multiply by 1000, you are not dealing with real multiplication. As a result you will start getting rollovers and hit negative values.

[daniel]

I'm guessing... no
but im not really that good at programming, so i probably am not right

[John]

Originally Posted by WingedPanther

Yes, under most languages. When you multiply by 1000, you are not dealing with real multiplication. As a result you will start getting rollovers and hit negative values.

Interesting! Can you elaborate more? What do you mean by "real" multiplication and "rollovers?"

[icepack]

you have an integer value

int
Occupies 32 bits or 4 bytes, which is:
-231 to 231-1 or -2,147,483,648 to 2,147,483,647.
Default value of 0.

what he's saying is, what happens when it gets past 2,147,483,647?
it starts off at the next number, negative 2 billion something.

imagine the number line as a circle and not a straight finite line


when i ran the loop, i printed the value after the loop ended.
here it is:

#include <iostream>
using namespace std;

int main()
{
signed int i=1;
while (i>0)
i=i*1000;
cout << i << endl;
}
-----output-------
[mjd0605@curly:~]$ g++ loop.cpp
[mjd0605@curly:~]$ ./a.out
-727379968
[mjd0605@curly:~]$

[WingedPanther]

Elaborating on what IcePack said:
you start at 1, then get 1000, then 1,000,000, then 1,000,000,000, then 1,000,000,000,000 which is outside the range supported by an integer. The high bits get thrown out and you get a completely different number, possibly interpretted as positive or negative (depending on the system). Keep this up for a while and you will eventually get to a negative number.

[John]

Very interesting, thanks

[kphilux]

That was a great response many thanks for the help.