Generally the operators that can't be overloaded are like that because overloading them could and probably would cause serious program errors or it is syntactically not possible,
For instance the sizeof operator returns the size of the object or type passed as an operand. It is evaluated by the compiler not at runtime so you can not overload it with your own runtime code. It is syntactically not possible to do. Even if it was pointer arithmetic relies on the correct value being returned by this operator since the compiler already knows how to calculate the correct value all overloading would do would be to allow you to calculate an incorrect value, something that would almost certainly lead to the program not working correctly.
Scope resolution and member access operators work on names rather than values. C++ has no syntax for writing code that works on names rather than values so syntactically these operators can not be overridden.
Again what useful purpose would overloading the conditional operator produce? I can think of none.
it cannot be operator overloaded.
There are 5 operators which cannot be overloaded. They are: * .* - class member access operator * :: - scope resolution operator * . - dot operator * ?:: - conditional operator * Sizeof() - operator Note:- This is possible only in C++.
The only "special" operators in C++ are those that cannot be overloaded. That is; the dot member operator (.), pointer to member operator (.*), ternary conditional operator (:?), scope resolution operator (::), sizeof() and typeof().
To access a hidden global variable, use the scope resolution operator ::
The if statementex.if (index < 5)printf("Index is less than 5\n");elseprintf("index is greater or equal to 5\n");(You can also replace the "if" with a "?" and the "else" with a "?" -- no, that would be syntax error)
it cannot be operator overloaded.
No.
There are 5 operators which cannot be overloaded. They are: * .* - class member access operator * :: - scope resolution operator * . - dot operator * ?:: - conditional operator * Sizeof() - operator Note:- This is possible only in C++.
1. Member-of operator (.) 2. Pointer-to-member-of operator (.*) 3. Ternary condition operator (?:) 4. Scope resolution operator (::) 5. sizeof operator 6. typeid operator
The only "special" operators in C++ are those that cannot be overloaded. That is; the dot member operator (.), pointer to member operator (.*), ternary conditional operator (:?), scope resolution operator (::), sizeof() and typeof().
conditional operator , size of operator , membership operator and scope resulation operator can not be overload in c++
:: operator can not be used in C.
The scope resolution operator, ::, overrides local scope and allows access to objects that are hidden due to global to local scope rules.
To access a hidden global variable, use the scope resolution operator ::
You use the scope resolution operator (::) whenever there is ambiguity as to which function or member you are referring to. For instance, if two functions in two separate namespaces have the same signature, you must use scope resolution to call the correct version of the function. Similarly, when calling a base class method from a derived overridden method, you must use scope resolution to ensure the base class method is called from the override.
The if statementex.if (index < 5)printf("Index is less than 5\n");elseprintf("index is greater or equal to 5\n");(You can also replace the "if" with a "?" and the "else" with a "?" -- no, that would be syntax error)
A hidden global variable must be one that has its scope blocked by a local variable of the same name. To access the hidden variable, use the scope resolution operator ::, such as is ::variable_name. If there is another reason for the hidden status, please clarify and restate the question.