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Question about ~ operator

Sudhakar Krishnamurthy
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Joined: Jun 02, 2003
Posts: 76
Can anyone shed some light on this behavior??

short i =0;
i>>=i; //compiles fine, an int is implicitly converted to a short here
i~=i; //throws a compiler error, explict casting is required here

Why?
TIA
Corey McGlone
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Joined: Dec 20, 2001
Posts: 3271
As far as I know, there is no such operator as ~=. ~ is the Unary NOT operator, which means that it only functions on one operand. The compound assignment operators are designed to be shorthand notations of Binary Operators, such as +, -, >>, etc.


SCJP Tipline, etc.
Sudhakar Krishnamurthy
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Joined: Jun 02, 2003
Posts: 76
i get it, thanks
Chris Allen
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Joined: Feb 01, 2003
Posts: 127
Originally posted by Sudhakar Krishnamurthy:
Can anyone shed some light on this behavior??

short i =0;
i>>=i; //compiles fine, an int is implicitly converted to a short here
i~=i; //throws a compiler error, explict casting is required here

Why?
TIA


I changed the code slightly as below:

The compiler indicates that it will not compile because of a possible loss of precision. This makes sense because you are trying to put an integer value (32 bits) into a short value (16 bits). Remember that the value of the short variable,i, will be promoted to an integer before applying the unary operator. Thus, an explicit cast is required. This can result in some strange results when attempting this operation.
 
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