How can an integer variable show a correct value even when the value crosses its range?

Question

I have executed this program on 32bit Windows 7 OS with Turbo C/C++ editor.

#include<stdio.h>
#include<conio.h>

void main()
{
    int a,b,c,d,e;
    clrscr();

    a = 25000;
    b = 10000;

    c = a + b;

    printf(" The value of c = %d\n", c );

    d = 5000;

    e = c - d;

    printf(" The value of e is %d\n", e );

    getch();
}

When I print c I get the value -30536, because the value 35000 crosses the max value. But when I use the same c in the expression 'e= c - d;' , I get the correct value as 30000. How is this possible?

David.. Please elaborate your answer...how can -30536 - 5000 be equal to 30000?? — Rohit S, Apr 20 '15 at 04:42
Overflow/lunderflow. BTW, if you really want to learn either C or C++, use an even halfway-modern compiler. — Ulrich Eckhardt, Apr 20 '15 at 04:43
You were happy with `25000 + 10000 = -30536`, so why not `-30536 - 5000 = 30000`? — David Heffernan, Apr 20 '15 at 04:43
What happens with `25000 + 10000` is the same with `-30536 - 5000` — Inder Kumar Rathore, Apr 20 '15 at 04:44
I have not stated that I was satisfied with '25000 + 10000 = -30536' ... I wanted to show what my system gave for '25000 + 10000 '!! When the c is not holding the value 35000 how can it give 30000 after deducting 5000 ?? That's my question. If it is really holding 35000 then why it is showing -30536. — Rohit S, Apr 20 '15 at 04:46
@RohitS it's very simple if you think number as binary not decimal. Don't user your brain to calculate decimal user CPU's brain to calculate binary — Inder Kumar Rathore, Apr 20 '15 at 04:48
It's not storing 35000. It's -30536. Back to basics. How can you accept `25000 + 10000 = -30536`? — David Heffernan, Apr 20 '15 at 04:48
Inder.. you mean to say that...25000 + 10000 = -30536 is correct??? If so ..how ?? — Rohit S, Apr 20 '15 at 04:50
You said it was fine. You said you understood why it happened. *When I print c I get the value -30536, because the value 35000 crosses the max value*. — David Heffernan, Apr 20 '15 at 04:51
In standard C your program has undefined behaviour. Your compiler is ancient and you should get a modern standard compliant compiler. And learn about undefined behaviour. — David Heffernan, Apr 20 '15 at 04:54
the main reason for the odd results is because the compiler you are using has a int size of 16bits, but modern hardware has an int size of either 32 or 64 bits and the underlying math is performed with the full register. — user3629249, Apr 20 '15 at 05:11
and [`void main()` is also wrong](http://stackoverflow.com/questions/204476/what-should-main-return-in-c-and-c?lq=1), because Turbo C existed before the first standard, it doesn't conform to the standard in many points, including `void main()` — phuclv, Apr 20 '15 at 07:52

Inder Kumar Rathore · Accepted Answer · 2015-04-20T05:06:18.400

4

Let's kick the decimal and our brain and use binary and CPU. The int is 16 bit here

Expression 25000 + 10000 (add the binary code very simple)

 01100001 10101000 = 25000
+00100111 00010000 = 10000
 10001000 10111000 = -30536

10001000 10111000 is the result which is negative 30536 as first bit is signed bit

Now expression -30536 - 5000 (-30536 + (-5000) )

 10001000 10111000 = -30536
+11101100 01111000 = -5000 
 01110101 00110000 = 30000

01110101 00110000 this result is positive 30000

Or -30536 - 5000 you can use 2's complement method to subtract. I won't explain it here.

edited Apr 20 '15 at 05:06

answered Apr 20 '15 at 04:54

Inder Kumar Rathore

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Both expressions are UB with 16 bit signed int. Surely you need to mention this. Or asker will think this can be relied upon. – David Heffernan Apr 20 '15 at 04:56
@DavidHeffernan I was editing this before your comment came. Anyways thanks – Inder Kumar Rathore Apr 20 '15 at 04:59
I can't see mention of the UB. Now Rohit is going to think he can rely on this. – David Heffernan Apr 20 '15 at 05:04
@DavidHeffernan can you elaborate abbreviation `UB`. Is it Under Bound? – Inder Kumar Rathore Apr 20 '15 at 05:08
@RohitS http://stackoverflow.com/questions/4240748/allowing-signed-integer-overflows-in-c-c http://stackoverflow.com/questions/16188263/is-signed-integer-overflow-still-undefined-behavior-in-c – phuclv Apr 20 '15 at 06:27

How can an integer variable show a correct value even when the value crosses its range?

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