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What are factors to be considered while overloading the operators?

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2011-01-31 06:18:06
2011-01-31 06:18:06

Java does not support operator Overloading.

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While overloading operators using member function it takes only one arguments(other pass implicitly)... but in case of friend fuction two parameters are required..


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Overloading will be done at compile time itself. Proof: If you try to narrow down the access modifier for a method from public in the parent class to private in the child class while overloading, the compiler will not let you do it.


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Maybe because Sun said so. We have to bear with so many other idiosyncrasies too. But I guess that comes with every language. There were two major reasons why operator overloading wasn't allowed in Java: "cleanliness" and compiler complexity. The main reason was the first, a personal preference choice made by Java's creator, James Gosling. Operator overloading, while useful, can be exceedingly confusing, much more so than method overloading. Given the human tendency to assign specific meanings to single symbols, it is hard to get programmers to wrap their heads around multiple meanings for operators. What this means is that there is a marked increase in programming errors when a language supports operator overloading. Since practically the same benefit can be obtained via methods, the Java designers decided that the increased programmer mistake rate was not worth supporting operator overloading. From a Java compiler (e.g. javac) design standpoint, supporting operator overloading is considerably more difficult than method overloading, requiring a more complex compiler.


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in C++ there is no real difference as operators are overloaded by implementing them as functions. However, while we differentiate between function overloads by the function signature (the number and type of parameters), operator overloads are distinguished only by the parameter types. The parameters are interpreted as operands, and the number of operands will depend upon whether the operator is unary, binary or ternary. That is, for any given operator, the number of operands will be the same for each overload you implement. The only exceptions are the unary increment (++) and decrement (--) operators as they each have postfix and prefix variants. In order to differentiate their signatures, an unreferenced or dummy parameter must be passed to the postfix variants.


There is no such thing as function overloading in C; that is a feature of C++. Function overloading allows us to provide two or more implementations of the same function. Typically, we use function overloading so that the same function can cater for different types. For instance, we might provide one implementation that is optimised to handle an integer argument while another is optimised to handle a real argument. We can also use function overloading to provide a common implementation of a function which can then be invoked by overloads that handle the low-level type conversions.


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The concept of Operator Overloading is similar to Method Overloading, in that the meaning of a given operator symbol changes according to the context it is being used in. That is, the semantics of the operator symbol are flexible, rather than fixed.The idea behind Operator Overloading is to take a common symbol, and adjust it's meaning to something logical for contexts other than what it was originally restricted to.The arithmetic operators ( + - * / ) are good examples. Using Operator Overloading, I could define that 'SomeArray + SomeValue' means that I should add SomeValue to the end of the array SomeArray.In general, Operator Overloading is what is called 'syntactic sugar' - it makes things more readable. For instance, the equivalent way to do the above example via method calls would be: SomeArray.addToEnd(SomeValue)The major problem with Operator Overloading is that it depends on people having the exact same interpretation of what an operator would mean in the new context, which is difficult to assure. Going back to the above example, there is some ambiguity as to where 'SomeArray + SomeValue' would mean to add in SomeValue - should SomeValue be added to the start of the array, or the end of the array? The answer is not obvious, and one would have to go look through the overload definition. While this confusion is also possible with methods, properly named methods (i.e. using addToEnd() rather than just add() ) helps avoid this entirely.For this reason, Java does not support user-defined Operator Overloading. Java does support certain operator overloading in narrow contexts, but only those defined by the language itself. That is, the '+' sign is overloaded to allow for string concatenation. However, the designer of Java (James Gosling) decided that his preference was to avoid Operator Overloading completely due to his perception of a "clean" language.


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The C bitwise operators allow programmers to operate upon the individual bits within a bit pattern of integral type. The bitwise shift operators (<< and >>) allow the programmer to shift bits left or right a specified number of binary positions while the bitwise NOT operator (~) allows the programmer to examine the ones-complement of an integral value. The bitwise AND (&), OR (|) and XOR (^) operators are binary operators (with two operands) and are useful for testing the state of individual bits within a bit pattern of integral type.


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I will not use operator overloading in C# to do anything. Operator overloading may lead to one operator has more than 1 semantic meaning. For example, we know 1 + 2 yields 3, and "1" + 2 yields "12". I do not like this overloading of the operator + being used for addition in Number hierarchy, while as the concatenation in strings. That is, one operator (+) has 2 significant semantics.And the question "find largest of two object" is too vague - what do you mean "largest"? and object? We know apple and orange are 2 objects, but how do you compare them, and find the largest one?????? (size, price or what???)


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