Dacă creăm doi sau mai mulți membri având același nume, dar diferiți ca număr sau tip de parametru, se numește supraîncărcare C++. În C++, putem supraîncărca:
- metode,
- constructori și
- proprietăți indexate
Se datorează faptului că acești membri au doar parametri.
Tipurile de supraîncărcare în C++ sunt:
- Supraîncărcarea funcției
- Supraîncărcarea operatorului
Supraîncărcarea funcției C++
Supraîncărcarea funcției este definită ca procesul de a avea două sau mai multe funcții cu același nume, dar diferiți în parametri este cunoscut ca supraîncărcarea funcției în C++. În supraîncărcarea funcției, funcția este redefinită folosind fie diferite tipuri de argumente, fie un număr diferit de argumente. Numai prin aceste diferențe, compilatorul poate diferenția între funcții.
șir ca matrice
The avantaj de supraîncărcare de funcții este că crește lizibilitatea programului, deoarece nu trebuie să utilizați nume diferite pentru aceeași acțiune.
Exemplu de supraîncărcare a funcției C++
Să vedem exemplul simplu de supraîncărcare a funcției în care schimbăm numărul de argumente ale metodei add().
// program de supraîncărcare a funcției când numărul de argumente variază.
#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,> Să vedem exemplul simplu când tipul argumentelor variază.
// Program de supraîncărcare a funcțiilor cu diferite tipuri de argumente.
#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<<> Supraîncărcarea funcției și ambiguitatea
Când compilatorul nu poate decide ce funcție va fi invocată printre funcțiile supraîncărcate, această situație este cunoscută ca supraîncărcarea funcției .
Când compilatorul arată eroarea de ambiguitate, compilatorul nu rulează programul.
Cauzele supraîncărcării funcțiilor:
- Conversie tip.
- Funcție cu argumente implicite.
- Funcție cu trecere prin referință.
- Conversie tip:
Să vedem un exemplu simplu.
#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<<> Unde tip de returnare este tipul de valoare returnat de funcție.
numele clasei este numele clasei.
operator op este o funcție de operator în care op este operatorul supraîncărcat, iar operatorul este cuvântul cheie.
Reguli pentru supraîncărcarea operatorului
- Operatorii existenți pot fi doar supraîncărcați, dar noii operatori nu pot fi supraîncărcați.
- Operatorul supraîncărcat conține cel puțin un operand de tipul de date definit de utilizator.
- Nu putem folosi funcția friend pentru a supraîncărca anumiți operatori. Cu toate acestea, funcția membru poate fi utilizată pentru a supraîncărca acești operatori.
- Când operatorii unari sunt supraîncărcați printr-o funcție membru, nu iau argumente explicite, dar, dacă sunt supraîncărcați de o funcție prietenă, iau un argument.
- Când operatorii binari sunt supraîncărcați printr-o funcție membru, ia un argument explicit, iar dacă sunt supraîncărcați printr-o funcție prietenă ia două argumente explicite.
Exemplu de supraîncărcare a operatorilor C++
Să vedem exemplul simplu de supraîncărcare a operatorului în C++. În acest exemplu, funcția operator void ++ () este definită (în interiorul clasei Test).
// program pentru supraîncărcarea operatorului unar ++.
#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> Să vedem un exemplu simplu de supraîncărcare a operatorilor binari.
// program de supraîncărcare a operatorilor binari.
#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>