Si creamos dos o más miembros con el mismo nombre pero diferentes en número o tipo de parámetro, se conoce como sobrecarga de C++. En C++, podemos sobrecargar:
- métodos,
- constructores y
- propiedades indexadas
Es porque estos miembros solo tienen parámetros.
Los tipos de sobrecarga en C++ son:
- Sobrecarga de funciones
- Sobrecarga del operador
Sobrecarga de funciones de C++
La sobrecarga de funciones se define como el proceso de tener dos o más funciones con el mismo nombre, pero con diferentes parámetros, lo que se conoce como sobrecarga de funciones en C++. En la sobrecarga de funciones, la función se redefine utilizando diferentes tipos de argumentos o un número diferente de argumentos. Sólo a través de estas diferencias el compilador puede diferenciar entre las funciones.
tabla de reacciones
El ventaja de la sobrecarga de funciones es que aumenta la legibilidad del programa porque no es necesario utilizar diferentes nombres para la misma acción.
Ejemplo de sobrecarga de funciones de C++
Veamos el ejemplo simple de sobrecarga de funciones donde cambiamos la cantidad de argumentos del método add().
// programa de sobrecarga de funciones cuando varía el número de argumentos.
#include using namespace std; class Cal { public: static int add(int a,int b){ return a + b; } static int add(int a, int b, int c) { return a + b + c; } }; int main(void) { Cal C; // class object declaration. cout<<c.add(10, 20)<<endl; cout<<c.add(12, 20, 23); return 0; } < pre> <p> <strong>Output:</strong> </p> <pre> 30 55 </pre> <p>Let's see the simple example when the type of the arguments vary.</p> <p>// Program of function overloading with different types of arguments.</p> <pre> #include using namespace std; int mul(int,int); float mul(float,int); int mul(int a,int b) { return a*b; } float mul(double x, int y) { return x*y; } int main() { int r1 = mul(6,7); float r2 = mul(0.2,3); std::cout << 'r1 is : ' <<r1<< std::endl; std::cout <<'r2 is : ' <<r2<< return 0; } < pre> <p> <strong>Output:</strong> </p> <pre> r1 is : 42 r2 is : 0.6 </pre> <h2>Function Overloading and Ambiguity</h2> <p>When the compiler is unable to decide which function is to be invoked among the overloaded function, this situation is known as <strong>function overloading</strong> .</p> <p>When the compiler shows the ambiguity error, the compiler does not run the program.</p> <p> <strong>Causes of Function Overloading:</strong> </p> <ul> <li>Type Conversion.</li> <li>Function with default arguments.</li> <li>Function with pass by reference.</li> </ul> <img src="//techcodeview.com/img/c-tutorial/89/c-overloading-function-2.webp" alt="C++ Overloading"> <ul> <li>Type Conversion:</li> </ul> <p> <strong>Let's see a simple example.</strong> </p> <pre> #include using namespace std; void fun(int); void fun(float); void fun(int i) { std::cout << 'Value of i is : ' < <i<< std::endl; } void fun(float j) { std::cout << 'value of j is : ' <<j<< int main() fun(12); fun(1.2); return 0; < pre> <p>The above example shows an error ' <strong>call of overloaded 'fun(double)' is ambiguous</strong> '. The fun(10) will call the first function. The fun(1.2) calls the second function according to our prediction. But, this does not refer to any function as in C++, all the floating point constants are treated as double not as a float. If we replace float to double, the program works. Therefore, this is a type conversion from float to double.</p> <ul> <li>Function with Default Arguments</li> </ul> <p> <strong>Let's see a simple example.</strong> </p> <pre> #include using namespace std; void fun(int); void fun(int,int); void fun(int i) { std::cout << 'Value of i is : ' < <i<< std::endl; } void fun(int a,int b="9)" { std::cout << 'value of a is : ' < <a<< <b<< int main() fun(12); return 0; pre> <p>The above example shows an error 'call of overloaded 'fun(int)' is ambiguous'. The fun(int a, int b=9) can be called in two ways: first is by calling the function with one argument, i.e., fun(12) and another way is calling the function with two arguments, i.e., fun(4,5). The fun(int i) function is invoked with one argument. Therefore, the compiler could not be able to select among fun(int i) and fun(int a,int b=9).</p> <ul> <li>Function with pass by reference</li> </ul> <p>Let's see a simple example.</p> <pre> #include using namespace std; void fun(int); void fun(int &); int main() { int a=10; fun(a); // error, which f()? return 0; } void fun(int x) { std::cout << 'Value of x is : ' <<x<< std::endl; } void fun(int &b) { std::cout << 'value of b is : ' < <b<< pre> <p>The above example shows an error ' <strong>call of overloaded 'fun(int&)' is ambiguous</strong> '. The first function takes one integer argument and the second function takes a reference parameter as an argument. In this case, the compiler does not know which function is needed by the user as there is no syntactical difference between the fun(int) and fun(int &).</p> <h2>C++ Operators Overloading</h2> <p>Operator overloading is a compile-time polymorphism in which the operator is overloaded to provide the special meaning to the user-defined data type. Operator overloading is used to overload or redefines most of the operators available in C++. It is used to perform the operation on the user-defined data type. For example, C++ provides the ability to add the variables of the user-defined data type that is applied to the built-in data types.</p> <p>The advantage of Operators overloading is to perform different operations on the same operand.</p> <p> <strong>Operator that cannot be overloaded are as follows:</strong> </p> <ul> <li>Scope operator (::)</li> <li>Sizeof</li> <li>member selector(.)</li> <li>member pointer selector(*)</li> <li>ternary operator(?:) </li> </ul> <h2>Syntax of Operator Overloading</h2> <pre> return_type class_name : : operator op(argument_list) { // body of the function. } </pre> <p>Where the <strong>return type</strong> is the type of value returned by the function. </p><p> <strong>class_name</strong> is the name of the class.</p> <p> <strong>operator op</strong> is an operator function where op is the operator being overloaded, and the operator is the keyword.</p> <h2>Rules for Operator Overloading</h2> <ul> <li>Existing operators can only be overloaded, but the new operators cannot be overloaded.</li> <li>The overloaded operator contains atleast one operand of the user-defined data type.</li> <li>We cannot use friend function to overload certain operators. However, the member function can be used to overload those operators.</li> <li>When unary operators are overloaded through a member function take no explicit arguments, but, if they are overloaded by a friend function, takes one argument.</li> <li>When binary operators are overloaded through a member function takes one explicit argument, and if they are overloaded through a friend function takes two explicit arguments. </li> </ul> <h2>C++ Operators Overloading Example</h2> <p>Let's see the simple example of operator overloading in C++. In this example, void operator ++ () operator function is defined (inside Test class).</p> <p>// program to overload the unary operator ++.</p> <pre> #include using namespace std; class Test { private: int num; public: Test(): num(8){} void operator ++() { num = num+2; } void Print() { cout<<'the count is: '<<num; } }; int main() { test tt; ++tt; calling of a function 'void operator ++()' tt.print(); return 0; < pre> <p> <strong>Output:</strong> </p> <pre> The Count is: 10 </pre> <p>Let's see a simple example of overloading the binary operators.</p> <p>// program to overload the binary operators.</p> <pre> #include using namespace std; class A { int x; public: A(){} A(int i) { x=i; } void operator+(A); void display(); }; void A :: operator+(A a) { int m = x+a.x; cout<<'the result of the addition two objects is : '<<m; } int main() { a a1(5); a2(4); a1+a2; return 0; < pre> <p> <strong>Output:</strong> </p> <pre> The result of the addition of two objects is : 9 </pre></'the></pre></'the></pre></x<<></pre></i<<></pre></i<<></pre></r1<<></pre></c.add(10,> Veamos el ejemplo sencillo cuando el tipo de argumentos varía.
// Programa de sobrecarga de funciones con diferentes tipos de argumentos.
#include using namespace std; int mul(int,int); float mul(float,int); int mul(int a,int b) { return a*b; } float mul(double x, int y) { return x*y; } int main() { int r1 = mul(6,7); float r2 = mul(0.2,3); std::cout << 'r1 is : ' <<r1<< std::endl; std::cout <<\'r2 is : \' <<r2<< return 0; } < pre> <p> <strong>Output:</strong> </p> <pre> r1 is : 42 r2 is : 0.6 </pre> <h2>Function Overloading and Ambiguity</h2> <p>When the compiler is unable to decide which function is to be invoked among the overloaded function, this situation is known as <strong>function overloading</strong> .</p> <p>When the compiler shows the ambiguity error, the compiler does not run the program.</p> <p> <strong>Causes of Function Overloading:</strong> </p> <ul> <li>Type Conversion.</li> <li>Function with default arguments.</li> <li>Function with pass by reference.</li> </ul> <img src="//techcodeview.com/img/c-tutorial/89/c-overloading-function-2.webp" alt="C++ Overloading"> <ul> <li>Type Conversion:</li> </ul> <p> <strong>Let's see a simple example.</strong> </p> <pre> #include using namespace std; void fun(int); void fun(float); void fun(int i) { std::cout << 'Value of i is : ' < <i<< std::endl; } void fun(float j) { std::cout << \'value of j is : \' <<j<< int main() fun(12); fun(1.2); return 0; < pre> <p>The above example shows an error ' <strong>call of overloaded 'fun(double)' is ambiguous</strong> '. The fun(10) will call the first function. The fun(1.2) calls the second function according to our prediction. But, this does not refer to any function as in C++, all the floating point constants are treated as double not as a float. If we replace float to double, the program works. Therefore, this is a type conversion from float to double.</p> <ul> <li>Function with Default Arguments</li> </ul> <p> <strong>Let's see a simple example.</strong> </p> <pre> #include using namespace std; void fun(int); void fun(int,int); void fun(int i) { std::cout << 'Value of i is : ' < <i<< std::endl; } void fun(int a,int b="9)" { std::cout << \'value of a is : \' < <a<< <b<< int main() fun(12); return 0; pre> <p>The above example shows an error 'call of overloaded 'fun(int)' is ambiguous'. The fun(int a, int b=9) can be called in two ways: first is by calling the function with one argument, i.e., fun(12) and another way is calling the function with two arguments, i.e., fun(4,5). The fun(int i) function is invoked with one argument. Therefore, the compiler could not be able to select among fun(int i) and fun(int a,int b=9).</p> <ul> <li>Function with pass by reference</li> </ul> <p>Let's see a simple example.</p> <pre> #include using namespace std; void fun(int); void fun(int &); int main() { int a=10; fun(a); // error, which f()? return 0; } void fun(int x) { std::cout << 'Value of x is : ' <<x<< std::endl; } void fun(int &b) { std::cout << \'value of b is : \' < <b<< pre> <p>The above example shows an error ' <strong>call of overloaded 'fun(int&)' is ambiguous</strong> '. The first function takes one integer argument and the second function takes a reference parameter as an argument. In this case, the compiler does not know which function is needed by the user as there is no syntactical difference between the fun(int) and fun(int &).</p> <h2>C++ Operators Overloading</h2> <p>Operator overloading is a compile-time polymorphism in which the operator is overloaded to provide the special meaning to the user-defined data type. Operator overloading is used to overload or redefines most of the operators available in C++. It is used to perform the operation on the user-defined data type. For example, C++ provides the ability to add the variables of the user-defined data type that is applied to the built-in data types.</p> <p>The advantage of Operators overloading is to perform different operations on the same operand.</p> <p> <strong>Operator that cannot be overloaded are as follows:</strong> </p> <ul> <li>Scope operator (::)</li> <li>Sizeof</li> <li>member selector(.)</li> <li>member pointer selector(*)</li> <li>ternary operator(?:) </li> </ul> <h2>Syntax of Operator Overloading</h2> <pre> return_type class_name : : operator op(argument_list) { // body of the function. } </pre> <p>Where the <strong>return type</strong> is the type of value returned by the function. </p><p> <strong>class_name</strong> is the name of the class.</p> <p> <strong>operator op</strong> is an operator function where op is the operator being overloaded, and the operator is the keyword.</p> <h2>Rules for Operator Overloading</h2> <ul> <li>Existing operators can only be overloaded, but the new operators cannot be overloaded.</li> <li>The overloaded operator contains atleast one operand of the user-defined data type.</li> <li>We cannot use friend function to overload certain operators. However, the member function can be used to overload those operators.</li> <li>When unary operators are overloaded through a member function take no explicit arguments, but, if they are overloaded by a friend function, takes one argument.</li> <li>When binary operators are overloaded through a member function takes one explicit argument, and if they are overloaded through a friend function takes two explicit arguments. </li> </ul> <h2>C++ Operators Overloading Example</h2> <p>Let's see the simple example of operator overloading in C++. In this example, void operator ++ () operator function is defined (inside Test class).</p> <p>// program to overload the unary operator ++.</p> <pre> #include using namespace std; class Test { private: int num; public: Test(): num(8){} void operator ++() { num = num+2; } void Print() { cout<<\'the count is: \'<<num; } }; int main() { test tt; ++tt; calling of a function \'void operator ++()\' tt.print(); return 0; < pre> <p> <strong>Output:</strong> </p> <pre> The Count is: 10 </pre> <p>Let's see a simple example of overloading the binary operators.</p> <p>// program to overload the binary operators.</p> <pre> #include using namespace std; class A { int x; public: A(){} A(int i) { x=i; } void operator+(A); void display(); }; void A :: operator+(A a) { int m = x+a.x; cout<<\'the result of the addition two objects is : \'<<m; } int main() { a a1(5); a2(4); a1+a2; return 0; < pre> <p> <strong>Output:</strong> </p> <pre> The result of the addition of two objects is : 9 </pre></\'the></pre></\'the></pre></x<<></pre></i<<></pre></i<<></pre></r1<<> Sobrecarga de funciones y ambigüedad
Cuando el compilador no puede decidir qué función se invocará entre las funciones sobrecargadas, esta situación se conoce como sobrecarga de funciones .
Cuando el compilador muestra el error de ambigüedad, el compilador no ejecuta el programa.
Causas de la sobrecarga de funciones:
- Conversión de tipo.
- Función con argumentos predeterminados.
- Función con pase por referencia.
- Conversión de tipo:
Veamos un ejemplo sencillo.
#include using namespace std; void fun(int); void fun(float); void fun(int i) { std::cout << 'Value of i is : ' < <i<< std::endl; } void fun(float j) { std::cout << \'value of j is : \' <<j<< int main() fun(12); fun(1.2); return 0; < pre> <p>The above example shows an error ' <strong>call of overloaded 'fun(double)' is ambiguous</strong> '. The fun(10) will call the first function. The fun(1.2) calls the second function according to our prediction. But, this does not refer to any function as in C++, all the floating point constants are treated as double not as a float. If we replace float to double, the program works. Therefore, this is a type conversion from float to double.</p> <ul> <li>Function with Default Arguments</li> </ul> <p> <strong>Let's see a simple example.</strong> </p> <pre> #include using namespace std; void fun(int); void fun(int,int); void fun(int i) { std::cout << 'Value of i is : ' < <i<< std::endl; } void fun(int a,int b="9)" { std::cout << \'value of a is : \' < <a<< <b<< int main() fun(12); return 0; pre> <p>The above example shows an error 'call of overloaded 'fun(int)' is ambiguous'. The fun(int a, int b=9) can be called in two ways: first is by calling the function with one argument, i.e., fun(12) and another way is calling the function with two arguments, i.e., fun(4,5). The fun(int i) function is invoked with one argument. Therefore, the compiler could not be able to select among fun(int i) and fun(int a,int b=9).</p> <ul> <li>Function with pass by reference</li> </ul> <p>Let's see a simple example.</p> <pre> #include using namespace std; void fun(int); void fun(int &); int main() { int a=10; fun(a); // error, which f()? return 0; } void fun(int x) { std::cout << 'Value of x is : ' <<x<< std::endl; } void fun(int &b) { std::cout << \'value of b is : \' < <b<< pre> <p>The above example shows an error ' <strong>call of overloaded 'fun(int&)' is ambiguous</strong> '. The first function takes one integer argument and the second function takes a reference parameter as an argument. In this case, the compiler does not know which function is needed by the user as there is no syntactical difference between the fun(int) and fun(int &).</p> <h2>C++ Operators Overloading</h2> <p>Operator overloading is a compile-time polymorphism in which the operator is overloaded to provide the special meaning to the user-defined data type. Operator overloading is used to overload or redefines most of the operators available in C++. It is used to perform the operation on the user-defined data type. For example, C++ provides the ability to add the variables of the user-defined data type that is applied to the built-in data types.</p> <p>The advantage of Operators overloading is to perform different operations on the same operand.</p> <p> <strong>Operator that cannot be overloaded are as follows:</strong> </p> <ul> <li>Scope operator (::)</li> <li>Sizeof</li> <li>member selector(.)</li> <li>member pointer selector(*)</li> <li>ternary operator(?:) </li> </ul> <h2>Syntax of Operator Overloading</h2> <pre> return_type class_name : : operator op(argument_list) { // body of the function. } </pre> <p>Where the <strong>return type</strong> is the type of value returned by the function. </p><p> <strong>class_name</strong> is the name of the class.</p> <p> <strong>operator op</strong> is an operator function where op is the operator being overloaded, and the operator is the keyword.</p> <h2>Rules for Operator Overloading</h2> <ul> <li>Existing operators can only be overloaded, but the new operators cannot be overloaded.</li> <li>The overloaded operator contains atleast one operand of the user-defined data type.</li> <li>We cannot use friend function to overload certain operators. However, the member function can be used to overload those operators.</li> <li>When unary operators are overloaded through a member function take no explicit arguments, but, if they are overloaded by a friend function, takes one argument.</li> <li>When binary operators are overloaded through a member function takes one explicit argument, and if they are overloaded through a friend function takes two explicit arguments. </li> </ul> <h2>C++ Operators Overloading Example</h2> <p>Let's see the simple example of operator overloading in C++. In this example, void operator ++ () operator function is defined (inside Test class).</p> <p>// program to overload the unary operator ++.</p> <pre> #include using namespace std; class Test { private: int num; public: Test(): num(8){} void operator ++() { num = num+2; } void Print() { cout<<\'the count is: \'<<num; } }; int main() { test tt; ++tt; calling of a function \'void operator ++()\' tt.print(); return 0; < pre> <p> <strong>Output:</strong> </p> <pre> The Count is: 10 </pre> <p>Let's see a simple example of overloading the binary operators.</p> <p>// program to overload the binary operators.</p> <pre> #include using namespace std; class A { int x; public: A(){} A(int i) { x=i; } void operator+(A); void display(); }; void A :: operator+(A a) { int m = x+a.x; cout<<\'the result of the addition two objects is : \'<<m; } int main() { a a1(5); a2(4); a1+a2; return 0; < pre> <p> <strong>Output:</strong> </p> <pre> The result of the addition of two objects is : 9 </pre></\'the></pre></\'the></pre></x<<></pre></i<<></pre></i<<> Donde el tipo de retorno es el tipo de valor devuelto por la función.
nombre de la clase es el nombre de la clase.
operación del operador es una función de operador donde op es el operador que se está sobrecargando y el operador es la palabra clave.
Reglas para la sobrecarga del operador
- Los operadores existentes sólo se pueden sobrecargar, pero los nuevos operadores no se pueden sobrecargar.
- El operador sobrecargado contiene al menos un operando del tipo de datos definido por el usuario.
- No podemos utilizar la función amiga para sobrecargar ciertos operadores. Sin embargo, la función miembro se puede utilizar para sobrecargar esos operadores.
- Cuando los operadores unarios se sobrecargan a través de una función miembro, no toman argumentos explícitos, pero, si están sobrecargados por una función amiga, toman un argumento.
- Cuando los operadores binarios se sobrecargan a través de una función miembro, toman un argumento explícito, y si están sobrecargados a través de una función amiga, toman dos argumentos explícitos.
Ejemplo de sobrecarga de operadores de C++
Veamos el ejemplo sencillo de sobrecarga de operadores en C++. En este ejemplo, la función del operador void operator ++ () está definida (dentro de la clase Test).
// programa para sobrecargar el operador unario ++.
#include using namespace std; class Test { private: int num; public: Test(): num(8){} void operator ++() { num = num+2; } void Print() { cout<<\\'the count is: \\'<<num; } }; int main() { test tt; ++tt; calling of a function \\'void operator ++()\\' tt.print(); return 0; < pre> <p> <strong>Output:</strong> </p> <pre> The Count is: 10 </pre> <p>Let's see a simple example of overloading the binary operators.</p> <p>// program to overload the binary operators.</p> <pre> #include using namespace std; class A { int x; public: A(){} A(int i) { x=i; } void operator+(A); void display(); }; void A :: operator+(A a) { int m = x+a.x; cout<<\\'the result of the addition two objects is : \\'<<m; } int main() { a a1(5); a2(4); a1+a2; return 0; < pre> <p> <strong>Output:</strong> </p> <pre> The result of the addition of two objects is : 9 </pre></\\'the></pre></\\'the> Veamos un ejemplo sencillo de sobrecarga de operadores binarios.
// programa para sobrecargar los operadores binarios.
#include using namespace std; class A { int x; public: A(){} A(int i) { x=i; } void operator+(A); void display(); }; void A :: operator+(A a) { int m = x+a.x; cout<<\\'the result of the addition two objects is : \\'<<m; } int main() { a a1(5); a2(4); a1+a2; return 0; < pre> <p> <strong>Output:</strong> </p> <pre> The result of the addition of two objects is : 9 </pre></\\'the>