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Question about ~ operator

 
Sudhakar Krishnamurthy
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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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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.
 
Sudhakar Krishnamurthy
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i get it, thanks
 
Chris Allen
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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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