INFO: Default Assignment Used, Not User-Defined operator=()Last reviewed: September 30, 1997Article ID: Q104650 |
The information in this article applies to:
SUMMARYThe Microsoft "C++ Language Reference" for C/C++ version 7.0 states the following:
...if the class declares a user-defined operator=() that takes an argument of type "reference to class-name", no default assignment operator is generated.(See the Memberwise Assignment and Initialization section of the Special Member Functions chapter.) This statement may be confusing. However, the term "argument" means formal parameter, not actual parameter.
MORE INFORMATIONThe following program outputs "Default assignment used". The default assignment operator is used for the assignment and not Derived::operator=() as might be expected. The Derived::operator(const Base &) function specifies a const Base & for a formal parameter. Because there is no operator=() that specifies a formal parameter of type "reference to class- name", the default assignment operator is generated and used. Some compiler vendors have interpreted the C++ language differently by using the term "argument" in the specification to mean actual parameter. In that case, the default assignment operator is not generated and the program above will output "Derived::operator=() called" because an object of type Derived is an object of type Base.
Sample Code
/* Compile Options needed: None */ #include <iostream.h> char* message1 = "Derived::operator=() called"; char* message2 = "Default assignment used"; class Base { public: char* OperatorCalled; Base() { OperatorCalled = message2; } }; class Derived : public Base { public: void operator=(const Base&) { OperatorCalled = message1; } Derived() {} }; void main() { Derived first, second; first = second; cout << first.OperatorCalled; } |
Additional query words: 8.00 8.00c 9.00 9.10
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