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Trying to understand Casting

 
Greenhorn
Posts: 11
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why does this result in -128?
int a = 128;
byte b = 0;
b =(byte) a;

why does this result in 44
int a = 300;
byte b = 0;
b =(byte) a;

I think I understand the fact that you wouldn't cast this way. Just trying to understand the output. If I am correct you would use casting when you have something like this
              int a = 50;
              byte b = 0;
              b = (byte) a * 2;
Although 50 * 2 = 100 which is in the range of byte, the use of the operator automatically converts the expression to an interger. Thanks for whatever you care share.


 
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Chad McAte wrote:why does this result in -128?
int a = 128;
byte b = 0;
b =(byte) a;



The bit pattern for a 128 value int is ... 0000 0000 0000 0000 0000 0000 1000 0000. When you explicitly cast it to a byte, even though this value doesn't fall into the range of the byte, the compiler will use the same bit pattern (for the lower 8 bits only).  And under twos complement rules, the bit pattern of 1000 0000, as a byte is -128.

Henry
 
Henry Wong
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Chad McAte wrote:
why does this result in 44
int a = 300;
byte b = 0;
b =(byte) a;



The bit pattern for a 300 value int is ... 0000 0000 0000 0000 0000 0001 0010 1100. When you explicitly cast it to a byte, even though this value doesn't fall into the range of the byte, the compiler will use the same bit pattern (for the lower 8 bits only).  And under twos complement rules, the bit pattern of 0010 1100, as a byte is 44.

Henry
 
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