3

I have started learning C# and need to clear some of confusion on overflown concept. As we aware of that if we exceed the limit of any data type , C# simply returns 0 .

eg : byte b = 255; if we increase the value of b by 1, then the value of b will be zero.for below code i am getting output as 256.

 using System;
 namespace HelloWorld{
     class program{
          static void Main(){
                  byte b = 255;
                  Console.WriteLine(b+1);
           }
       }
  }

Instead of 0, i am getting output as 256 , which is out of the limit of b of type byte. How is this possible ?.

using System;
namespace HelloWorld{
class program{
    static void Main(){
        byte b = 255;
        b = b+1
        Console.WriteLine(b);
    }
}

for above code i am getting compilation error I.e error CS0266: Cannot implicitly convert type int' tobyte'. An explicit conversion exists (are you missing a cast?)

Help !!!!

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3
  • 3
    “As we aware of that if we exceed the limit of any data type , C# simply returns 0.” No, it wraps. And addition on a byte always produces at least an int, for reasons (but in this case you’re adding an int literal – 1 – anyway, so it makes sense). Commented Jul 30, 2018 at 5:44
  • 2
    byte b = 255;b++; gives you the expected result Commented Jul 30, 2018 at 5:47
  • There simply is no operator + defined for byte. So the compiler automatically uses the one for int. The same applies to short, char and probably a few other integral types. Commented Jul 30, 2018 at 6:39

3 Answers 3

Reset to default
8

The (clue to the) answer to your queries about the compilation error and 256 is in the output of the following snippet:

byte b = 255;
Console.WriteLine($"{b + 1}, {(b+1).GetType()}");
Console.ReadLine();

Output:

256, System.Int32 <-- not `byte`

As for wrapping, I think the following snippet illustrates the concept quite well.

for (int i = 250; i < 260; i++)
{
     byte b = (byte)i;
       Console.WriteLine($"{i} => {b,3} ({Convert.ToString(b, 2).PadLeft(8, '0')})");
}

Output:

250 => 250 (11111010)
251 => 251 (11111011)
252 => 252 (11111100)
253 => 253 (11111101)
254 => 254 (11111110)
255 => 255 (11111111)
256 =>   0 (00000000)
257 =>   1 (00000001)
258 =>   2 (00000010)
259 =>   3 (00000011)
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Comments

1 
byte b = 255;
b = b + 1;

This throws an error because b is of type byte and it takes 1 as int literal. You need to do type casting here. 

 byte b = 255;
 b++;
 Console.WriteLine(b);

This will gives you 0.

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2 Comments

There is no type casting shown here. you need to convert 1 into byte that is 'b += Convert.ToByte(1);'
Just b += 1; would be fine. There are subtle differences between that and b = b + 1; in terms of conversions.
0

It is a little bit different here. In your first example, the expression b+1 gets converted to an integer, as 1 by it self is always an integer, that's why your output is 256.

In the second example, you are trying to at an integer to an byte again, which is converted again to an integer, and then you want to assign the result back to an byte, which is not possible without a cast.

So, in c#, when adding numbers, the result is automatically converted to the type that can hold the larger value.

Same goes for double or float.

float f = 0.1;
double d = 0.1;
var x = f + d; // x is a double

To avoid the automatic conversion, you need to cast like b + (byte)1

To avoid the automatic conversion, you need to cast like (byte)(b + 1) (thanks @fubo)

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1 Comment

It should be (byte)(b+1)

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