Why doesn't this overflow?

Question

Given this code:

int x = 20000;
int y = 20000;
int z = 40000;

// Why is it printing WTF? Isn't 40,000 > 32,767?
if ((x + y) == z) Console.WriteLine("WTF?");

And knowing an int can hold −32,768 to +32,767. Why doesn't this cause an overflow?

Did you try printing out "x+y", then printing out z? Maybe they both get stored as the same negative number. — Tyler
– Tyler, Commented Jul 14, 2009 at 5:41
I wish all the people who posted later than the original correct answer would delete their redundant answers... — GManNickG
– GManNickG, Commented Jul 14, 2009 at 6:02

Blixt · Accepted Answer · 2009-07-14 06:22:51Z

21

In C#, the int type is mapped to the Int32 type, which is always 32-bits, signed.

Even if you use short, it still won't overflow because short + short returns an int by default. If you cast this int to short - (short)(x + y) - you'll get an overflowed value. You won't get an exception though. You can use checked behavior to get an exception:

using System;

namespace TestOverflow
{
    class Program
    {
        static void Main(string[] args)
        {
            short x = 20000;
            short y = 20000;
            short z;

            Console.WriteLine("Overflowing with default behavior...");
            z = (short)(x + y);
            Console.WriteLine("Okay! Value is {0}. Press any key to overflow " +
                "with 'checked' keyword.", z);
            Console.ReadKey(true);

            z = checked((short)(x + y));
        }
    }
}

You can find information about checked (and unchecked) on MSDN. It basically boils down to performance, because checking for overflow is a little bit slower than ignoring it (and that's why the default behavior is usually unchecked, but I bet that in some compilers/configurations you'll get an exception on the first z assignment.)

edited Jul 14, 2009 at 6:22

answered Jul 14, 2009 at 5:59

Blixt

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2 Comments

Vinko Vrsalovic Over a year ago

+1 I'd just add Scoregraphic's link here to make it a complete and great answer.

Blixt Over a year ago

Thanks for the tip, I added some links to Int32, int and short types =)

Scoregraphic · Accepted Answer · 2009-07-14 05:41:14Z

10

http://msdn.microsoft.com/en-us/library/5kzh1b5w.aspx

Type: int Range: -2,147,483,648 to 2,147,483,647

answered Jul 14, 2009 at 5:41

Scoregraphic

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2 Comments

jasonh Over a year ago

Downvoted because the answer isn't complete. See the answer that describes checked arithmetic.

Scoregraphic Over a year ago

The answer is complete in regard of the original question (why is there no overflow). So in my opinion, a downvote is wrong. But do an upvote on your prefered solution instead.

Falaina · Accepted Answer · 2009-07-14 06:02:48Z

6

While everyone is correct in saying that an "int" type on a 32 bit machine is most likely 2^32, there is a glaring flaw in your methodology.

Let's assume that int was 16 bit. You're assigning a value that will overflow z, so z itself is overflowed. When you calculate x+y you're also overflowing the int type, it's very likely that both cases will overflow to the same value, meaning you'd hit your equality regardless(this is probably compiler dependent, I'm not quite sure whether x+y will be promoted).

The correct way to do your experiment would be for z to have a larger data type than x and y. For example(Sorry for plain C, I'm not much of C# person. Hopefully it illustrated the methodology, however.)

int x = INT_MAX;
int y = INT_MAX;
int sum = x + y;
long long z = INT_MAX+INT_MAX;
if(sum == z) 
 printf("Why didn't sum overflow?!\n");

Comparing sum and z is important as comparing x+y and z may still come out fine depending on how the compiler handles promotion.

edited Jul 14, 2009 at 6:02

answered Jul 14, 2009 at 5:54

Falaina

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1 Comment

Kredns Over a year ago

Great C example. I actually got this off of a C guy's blog, but asked it in C# because were on SO and it gave the same result.

Jeff S · Accepted Answer · 2009-07-14 05:41:36Z

2

In C#, an Int is 4 bytes. So it maxes out at 2^31 or 2,147,483,648. If you want a 2 byte integer, use a short instead of an int.

answered Jul 14, 2009 at 5:41

Jeff S

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Comments

score 2 · Accepted Answer · 2009-07-14 05:57:28Z

2

Because an int in .NET is a signed 32 bit number with a range of -2,147,483,648 to 2,147,483,647.

Reference : http://msdn.microsoft.com/en-us/library/5kzh1b5w(VS.80).aspx

edited Jul 14, 2009 at 5:57

answered Jul 14, 2009 at 5:42

user10644

Comments

Vinay Sajip · Accepted Answer · 2009-07-14 05:41:16Z

1

Because ints are 32-bit, holding values up to ±2GB.

answered Jul 14, 2009 at 5:41

Vinay Sajip

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Gábor Hargitai · Accepted Answer · 2009-07-14 05:41:44Z

1

An int's size is 4 byte so it can hold at least 2^31 which is around 2 billion.

answered Jul 14, 2009 at 5:41

Gábor Hargitai

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Christian C. Salvadó · Accepted Answer · 2009-07-14 05:42:00Z

1

The int keyword maps to the .NET Framework Int32 type, which can hold integers in the range from -2,147,483,648 to 2,147,483,647.

answered Jul 14, 2009 at 5:42

Christian C. Salvadó

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Comments

this. __curious_geek · Accepted Answer · 2009-07-14 05:42:29Z

1

in C# int (System.Int32) is of 32 bits which can happily store this value.

answered Jul 14, 2009 at 5:42

this. __curious_geek

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Comments

anishMarokey · Accepted Answer · 2009-07-14 05:48:35Z

1

you are given to print the result as "WTF?" .Then how should it display other value.

Int means int32 its range is –2147483648 to 2147483647 you are given the range of int16 :–32768 to 32767

This is the reason it is not throwing any error

answered Jul 14, 2009 at 5:48

anishMarokey

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Comments

S M Kamran · Accepted Answer · 2009-07-14 06:26:56Z

First of all your code is in the range for int... However if it were not in the range then it wont complain either... coz you are never assigning a value back to any variable after doing X+Y in your if check...

Suppose if you were doing X * Y then it'll be calculated and the result would be a long value then the value from variable Z is taken and promoted to a long then both would be compared... Remember the casting from a lower range primitive to upper range primitive value is implicit.

int x = 200000; //In your code it was 20000
int y = 200000; //In your code it was 20000
int z = 40000;

// Why is it printing WTF? Isn't 40,000 > 32,767?
// Note: X + Y = 200000 and not < 32,767 
// would pass compiler coz you are not assigning and values are compared as longs
// And since it's not equals to 40,000 the WTF did not got printed
if ((x + y) == z) Console.WriteLine("WTF?");     
// And x * y >= z is true WTF MULTIPLY got printed
if ((x * y) >= z) Console.WriteLine("WTF MULTIPLY?"); 
// Compiler would fail since x can't hold 40,00,00,00,000
x = x * y;

BenMorel · Accepted Answer · 2014-05-24 23:37:45Z

0

All the above is true, however, it's important to know, that if you assign a number greater than 2^32, it will not compile!

edited May 24, 2014 at 23:37

BenMorel

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answered Jul 14, 2009 at 7:44

Clangon

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