Method overloading is widely used feature of most of the programming languages by why based on only class names and method argument lists? No method overloading is not possible in case of different return type, because compiler can't figure that which method he need to call.. For most programming languages that support method overloading (Java, C#, C++, ), if the parameter types are different, then the return types can also be different. You can overload on return type, but only for a type conversion operator function. When we call overloaded functions, the decision of which function is to be executed is known as function overloading resolution. 49, d3 = d1 + d2; here, d1 calls the operator function of its class object and takes d2 as a parameter, by which operator function return object and the result will reflect in the d3 object. And generally C++ expressions are context independent, which makes things easier and more reliable. Data types of the parameters 2.4. 26, Distance operator+(Distance &d2), here return type of function is distance and it uses call by references to pass an argument. In particular, for any type T, By using our site, you
You can not overload the function by differing only their return type. Only the const and volatile type-specifiers at the outermost level of the parameter type specification are ignored in this fashion; const and volatile type-specifiers buried within a parameter type specification are significant and can be used to distinguish overloaded function declarations. This modified text is an extract of the original Stack Overflow Documentation created by following contributors and released under CC … If the signatures are not same, then they can be overloaded. Function overloading is used for code reusability and also to save memory. Code maintenance is easy. Hiding of all overloaded methods with same name in base class, Write one line functions for strcat() and strcmp(), Functions that are executed before and after main() in C, Forward List in C++ | Set 2 (Manipulating Functions), List in C++ | Set 2 (Some Useful Functions), strtok() and strtok_r() functions in C with examples, Inbuilt library functions for user Input | scanf, fscanf, sscanf, scanf_s, fscanf_s, sscanf_s, strdup() and strndup() functions in C/C++, Left Shift and Right Shift Operators in C/C++, Map in C++ Standard Template Library (STL), Initialize a vector in C++ (5 different ways), Write Interview
at this point just having different names for the 2 functions feels a lot more K.I.S.S. This code shows one way to use the C++ overload rule using the function return type instead of its signature. Const or volatile are only used as a basis for overloading if they are used in a class to apply to the this pointer for the class, not the function's return type. className operator++(int); Which of the following is the general syntax of the function prototype to overload the post-increment operator as a member function? C didn't even have function overloading*. For example, following program fails in compilation with error “redefinition of `int f(int)’ “. The reason you can't do it for ordinary functions is that a call expression that currently on its own resolves to a specific overload, then generally wouldn't. no, the return types may be different but you cannot overload only on different return types - the overloaded functions must differ in at least one of the parameter types Jan 22 '11 # 2 Notice that a function cannot be overloaded only by its return type. Parameters that differ only in pointer type declaration vs array type declaration are considered to be equivalent and are not considered in function overloading. 6) Two parameter declarations that differ only in their default arguments are equivalent. Why not distinguish between methods based on their return values? error: functions that differ only in their return type cannot be overloaded void DLIB_FORTRAN_ID(dgesvd) (const char* jobu, const char* jobvt, Building from Xcode seems to work, I think this is related to this issue: #16 Function overloading and return type. Unfortunately, this doesn't work well with auto as it returns a Temp object by default. void area int area (); // error In C++, following function declarations cannot be overloaded. For instance, C++ permits us to add two variables of user-defined types with, the same syntax that is applied to the basic types. Only the second and subsequent array dimensions are significant in parameter types. Note that this list applies only to explicitly declared functions and those that have been introduced through using declarations: . At least one of its parameters must have a different type. We cannot overload functions on the basis of return type. At least one of its parameters must have a different type. Function which differs only in return type also cannot be overloaded. Operators can be overloaded either for objects of the user-defined types, or for a combination of objects of the user-defined type and objects of the built-in type. A function can be overloaded with a different return type if it has all the parameters same.