Because certain operators, such as new and sizeof must provide predictable results. Overloading them would achieve nothing except to introduce unpredictability, thus overloading is not permitted.
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 ternary operator (known as the conditional operator in C++) cannot be overloaded because it is impossible to pass a test operand and two expression operands (either or both of which may be comma-separated) to a function. You can only pass values or references as arguments to a function. Even if it were possible, built-in functions and operators that rely on the conditional operator would likely break. Like all the other operators that cannot be overloaded (sizeof, typeid, ::, . and .*) the results must always be predictable because built-in operators and functions rely on them so heavily.
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().
1. Member-of operator (.) 2. Pointer-to-member-of operator (.*) 3. Ternary condition operator (?:) 4. Scope resolution operator (::) 5. sizeof operator 6. typeid operator
it cannot be operator overloaded.
comma (,) is an example
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)
Yes. Any function can be overloaded. However you cannot override a static member function. Only instance members can be overridden.
You cannot create any new operators in C++. You can only overload the existing ones (although some, such as sizeof, new and delete cannot be overloaded). The only way to create a new operator is to implement it as a standard function with a named identifier. For instance, sqrt() is the standard library function that provides the square root operator, for which no real operator exists.
The C standard library IO facilities are not extensible. For instance, the printf() and scanf() functions cannot handle user-defined types. However, the C++ standard library provides IO streams with insertion and extraction operators (<< and >>) that can be overloaded to support any user-defined type.
5k 7 3k is a string of terms: it is not an equation (or inequality) and so cannot have an answer. It is not even an expression since there are no operators between the three terms.5k 7 3k is a string of terms: it is not an equation (or inequality) and so cannot have an answer. It is not even an expression since there are no operators between the three terms.5k 7 3k is a string of terms: it is not an equation (or inequality) and so cannot have an answer. It is not even an expression since there are no operators between the three terms.5k 7 3k is a string of terms: it is not an equation (or inequality) and so cannot have an answer. It is not even an expression since there are no operators between the three terms.
A C operator is not a job or profession but rather a coding language. C operators perform certain tasks in programming such as a "+ " operator performs addition.