Adding long numbers gives me minus result?

Question

when i try to add two long numbers it gives me minus result :

#include<iostream>
using namespace std;
int main ()
{
 int a=1825228665;
 int b=1452556585;
 cout<<a+b;
 return 0;
}

This gives me:

-1017182046

integer overflow, you exceed the max value for integer (specific to your compiler/os) — EdChum, Apr 10 '14 at 12:44
That's an overflow error. The result of the addition is too big to fit in a `int` variable. Even with an `unsigned int` you'd still get garbage. You should use a larger type, eg `long` or `long long`. — Panagiotis Kanavos, Apr 10 '14 at 12:45
@PanagiotisKanavos An unsigned integer will work fine. The result is approximately 3.2 billion, well within the range of an unsigned 32-bit integer. — Some programmer dude, Apr 10 '14 at 12:46
@JoachimPileborg I don't thing the question has to do with the specific values. `int`s are small and it's easy to get an overflow error. Besides, `int` isn't guaranteed to be 32 bit. The standard specifies minimum sizes: 16-bit for int, 32-bit for long — Panagiotis Kanavos, Apr 10 '14 at 12:47
When i try to add ULL to the number it work well , but i want to identify the variable a as ULL even if it doesn't equal a known value — user3516293, Apr 10 '14 at 12:49
You don't need a 64 bit int for this....but you can declare an `unsigned long long int` on some systems — David Wilkins, Apr 10 '14 at 12:50
@DavidWilkins i'm not talking specifically about this numbers , when i try to write unsigned long long int a it don't work ! — user3516293, Apr 10 '14 at 12:52
This has devolved into a discussion and is no longer an answerable question — David Wilkins, Apr 10 '14 at 12:57

Ardel · Answer 1 · 2014-04-10T12:52:36.240

0

It's overflowing of the type. When you add two big number that the result can't be stored in chosen type it get overfloved. In most cases it will wraped the number, but it's not defined in the standard. So in some compilers the result is undefined.

For int and other numeric types when program can't store this big number in it we can see a overflow of it.

Lets say that int can store number from -10 to 10, when you do this:

int a = 10;
int b = a+1;

You will get -10 in b or some random value (it can be anything because the result is undefined)

edited Apr 10 '14 at 12:52

answered Apr 10 '14 at 12:47

Ardel

315
2
9

1

I tried its giving 11. – Himanshu Apr 10 '14 at 12:49
1

Re your first statement: when an arithmetic operation results in a value which can't be represented in the type, the behavior is undefined. Most implementations do wrap, but it's not at all required. – James Kanze Apr 10 '14 at 12:50
Ok James Kanze,thanks. You're right – Ardel Apr 10 '14 at 12:50
@Himanshu That's because your `int` can probably store more than just `[-10, 10]`. – James Kanze Apr 10 '14 at 12:50

marcyb5st · Answer 2 · 2014-04-10T12:55:28.883

0

That's because the results overflows. Since the first bit in numeric signed data types is used for the sign representation. The specific representation is called Two's complement (Wikipedia article here). Practically a 1 in this bit maps to a - while a 0 to +. The solution to this problem is using a larger data type like long. Larger it means that the memory used to store it is bigger so the range of values increases.

edited Apr 10 '14 at 12:55

answered Apr 10 '14 at 12:50

marcyb5st

252
1
5

i used long but didn't work too , how can i declare a variable as ULL ? I tried unsigned long long int a but it didn't work. – user3516293 Apr 10 '14 at 12:54
Ask that as a different question please – David Wilkins Apr 10 '14 at 12:57

Adding long numbers gives me minus result?

2 Answers2