Jei sukuriame du ar daugiau narių, turinčių tą patį pavadinimą, bet skirtingą skaičiumi ar parametro tipu, tai vadinama C++ perkrovimu. C++ kalboje galime perkrauti:
- metodai,
- konstruktoriai ir
- indeksuotos savybės
Taip yra todėl, kad šie nariai turi tik parametrus.
C++ perkrovos tipai yra šie:
- Funkcijų perkrovimas
- Operatoriaus perkrova
C++ funkcijų perkrovimas
Funkcijų perkrovimas apibrėžiamas kaip dviejų ar daugiau funkcijų su tuo pačiu pavadinimu, tačiau skirtingų parametrų, procesas, vadinamas funkcijų perkrovimu C++. Perkraunant funkciją, funkcija iš naujo apibrėžiama naudojant skirtingų tipų argumentus arba skirtingą argumentų skaičių. Tik pagal šiuos skirtumus kompiliatorius gali atskirti funkcijas.
cout
The pranašumas Funkcijų perkrovimas padidina programos skaitomumą, nes tam pačiam veiksmui nereikia naudoti skirtingų pavadinimų.
C++ funkcijų perkrovos pavyzdys
Pažiūrėkime paprastą funkcijų perkrovimo pavyzdį, kai keičiame metodo add() argumentų skaičių.
// funkcijos perkrovimo programa, kai skiriasi argumentų skaičius.
#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,> Pažiūrėkime paprastą pavyzdį, kai skiriasi argumentų tipai.
// Funkcijos perkrovimo įvairių tipų argumentais programa.
#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<<> Funkcijų perkrovimas ir dviprasmiškumas
Kai kompiliatorius negali nuspręsti, kuri funkcija turi būti iškviesta tarp perkrautų funkcijų, ši situacija vadinama funkcijų perkrova .
Kai kompiliatorius parodo dviprasmiškumo klaidą, kompiliatorius nepaleidžia programos.
Funkcijos perkrovos priežastys:
- Tipas Konversija.
- Funkcija su numatytaisiais argumentais.
- Funkcija su praėjimu pagal nuorodą.
- Tipo konversija:
Pažiūrėkime paprastą pavyzdį.
#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<<> Kur grąžinimo tipas yra funkcijos grąžintos reikšmės tipas.
klasės_pavadinimas yra klasės pavadinimas.
operatorius op yra operatoriaus funkcija, kur op yra perkrautas operatorius, o operatorius yra raktinis žodis.
Operatoriaus perkrovos taisyklės
- Esami operatoriai gali būti tik perkrauti, bet nauji operatoriai negali būti perkrauti.
- Perkrautas operatorius turi bent vieną vartotojo apibrėžto duomenų tipo operandą.
- Negalime naudoti draugo funkcijos tam tikrų operatorių perkrovimui. Tačiau nario funkcija gali būti naudojama tiems operatoriams perkrauti.
- Kai unariniai operatoriai perkraunami per nario funkciją, nepriima jokių aiškių argumentų, tačiau, jei juos perkrauna draugo funkcija, imamas vienas argumentas.
- Kai dvejetainiai operatoriai yra perkraunami per nario funkciją, imamas vienas aiškus argumentas, o jei jie yra perkrauti per draugo funkciją, - du aiškūs argumentai.
C++ operatorių perkrovos pavyzdys
Pažiūrėkime paprastą operatoriaus perkrovimo C++ pavyzdį. Šiame pavyzdyje apibrėžta negaliojančio operatoriaus ++ () operatoriaus funkcija (testo klasėje).
// programa, skirta perkrauti unarinį operatorių ++.
#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> Pažiūrėkime paprastą dvejetainių operatorių perkrovimo pavyzdį.
// programa, skirta perkrauti dvejetainius operatorius.
#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>