YES only if: The return type in the child class is a sub-type of the return type of the parent class.
Ex:
public int getName() {} - Parent class method
public String getName() {} - Child class method
If we implement the above two methods in the class hierarchy we will get compilation errors. whereas if we try the below implementation it will work:
public Object getName() {} - Parent class method
public String getName() {} - Child class method
This is because String is a sub-type of Object and hence it works.
This is also called as Covariant Overriding.
Note: This feature is available only from Java 1.5. Earlier versions of Java expect overriding methods to have exactly the same return type as the super class method.
compiler would not know the information of return type at compile time
but it finds only at runtime.
No. Function overloading only pertains to the type of parameters.
Yes, but overloaded functions cannot differ by return type alone. The function signature must also differ in some way.
No
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.
A function must include an interface and an implementation. In some programming languages the two may be declared separately, particularly if the language is declarative. The interface typically consists of the function's return type followed by the function name and the type of its arguments, if any, usually enclosed in parenthesis (often round brackets). In untyped languages, the return type and the type of arguments may be optional, but the arguments must be formally named while the return value usually has the same name as the function. In typed languages the types must be specified but the names are optional unless the interface and implementation are combined. In typed languages that support function overloading, the function name and the number and type of the arguments form the function's unambiguous signature. The signature also includes the constness of the function and its arguments where supported, but does not include the return type nor any argument default values that may be provided. The function's implementation must duplicate the interface (if declared separately) but must also formally name the arguments. The function body is usually parenthesised (often in curly braces).
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.
The need for function overloading is to allow the same function call to accept different parameter types and/or a different number of parameters. The function signature differentiates the functions. The return type is also part of the signature, and needn't be the same for every overload. Where the function signature is largely the same, with the same count of parameters but a different parameter type and return type, template functions can provide an alternative method of creating the function overloads. The compiler generates the necessary code for you, on an as-required basis, so you only need to write the function once, rather than once for each type.
To overload a method in java, you only needs to write a method with the same name of other in your class but with diferente kind or number of parameters. For example: class MyClass{ void printNumber(int x){ System.out.println(x); } void printNumber(int x, int y){ System.out.println(x+y); } } the type of the value that your method return could change if you want.
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.
Both A function and a Sub carry out a procedure, but only A function returns a result while a Sub does not return a result.
A function must include an interface and an implementation. In some programming languages the two may be declared separately, particularly if the language is declarative. The interface typically consists of the function's return type followed by the function name and the type of its arguments, if any, usually enclosed in parenthesis (often round brackets). In untyped languages, the return type and the type of arguments may be optional, but the arguments must be formally named while the return value usually has the same name as the function. In typed languages the types must be specified but the names are optional unless the interface and implementation are combined. In typed languages that support function overloading, the function name and the number and type of the arguments form the function's unambiguous signature. The signature also includes the constness of the function and its arguments where supported, but does not include the return type nor any argument default values that may be provided. The function's implementation must duplicate the interface (if declared separately) but must also formally name the arguments. The function body is usually parenthesised (often in curly braces).
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.
The need for function overloading is to allow the same function call to accept different parameter types and/or a different number of parameters. The function signature differentiates the functions. The return type is also part of the signature, and needn't be the same for every overload. Where the function signature is largely the same, with the same count of parameters but a different parameter type and return type, template functions can provide an alternative method of creating the function overloads. The compiler generates the necessary code for you, on an as-required basis, so you only need to write the function once, rather than once for each type.
To overload a method in java, you only needs to write a method with the same name of other in your class but with diferente kind or number of parameters. For example: class MyClass{ void printNumber(int x){ System.out.println(x); } void printNumber(int x, int y){ System.out.println(x+y); } } the type of the value that your method return could change if you want.
When you overload the -- and ++ operators in C++, if you want the pre version, aka --var, simply declare the function without an argument; whereas if you want the post version, aka var--, simply declare the function with an argument of type int. class abc { public: abc& operator++(); // pre-increment form abc& operator++(int); // post-increment form ... }; abc::operator++() { ... pre increment stuff return *this; } abc::operator++(int) { ... post increment stuff return *this; }
The fork() library function makes a copy of the calling process. The original copy continues with a return value of the process ID of the new process, while the new copy continues with a return value of zero. If there is an error, the original process continues with a return value of -1. So, the answer is that fork is a process creation function.
Caller and callee relate to function calls. The caller is the code point that made the call to a function while the function is the callee. The callee returns control to the caller via the return address that was pushed onto the stack by the caller. void foo() {} int main() { foo(); } In the minimal example above, the main function is the caller while the foo function is the callee.
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.
If you live in your Grand-Daughters home that she pays for and if she supplies more than half of your support. Under these circumstances, you would be considered as her dependent on her tax return.
function factorial(n) { var x=1; while(n>1) x*=(n--); return x; }