이름은 같지만 매개변수의 개수나 유형이 다른 멤버를 두 개 이상 만드는 경우를 C++ 오버로딩이라고 합니다. C++에서는 다음을 오버로드할 수 있습니다.
- 행동 양식,
- 생성자 및
- 인덱싱된 속성
이 멤버에는 매개변수만 있기 때문입니다.
C++의 오버로딩 유형은 다음과 같습니다.
- 함수 오버로딩
- 연산자 오버로딩
C++ 함수 오버로딩
함수 오버로딩은 이름은 같지만 매개변수가 다른 두 개 이상의 함수를 갖는 프로세스로 정의됩니다. C++에서는 함수 오버로딩이라고 합니다. 함수 오버로딩에서는 다른 유형의 인수나 다른 개수의 인수를 사용하여 함수를 재정의합니다. 컴파일러는 이러한 차이점을 통해서만 함수를 구별할 수 있습니다.
100개 중 25개
그만큼 이점 함수 오버로딩의 장점은 동일한 작업에 대해 다른 이름을 사용할 필요가 없기 때문에 프로그램의 가독성이 높아진다는 것입니다.
C++ 함수 오버로딩 예
add() 메소드의 인수 수를 변경하는 함수 오버로딩의 간단한 예를 살펴보겠습니다.
// 인수 개수가 다를 때 함수 오버로딩을 수행하는 프로그램입니다.
#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,> 인수 유형이 다양한 경우의 간단한 예를 살펴보겠습니다.
// 다양한 유형의 인수를 사용하여 함수 오버로딩을 수행하는 프로그램입니다.
#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<<> 함수 오버로딩과 모호성
컴파일러가 오버로드된 함수 중에서 어떤 함수를 호출할지 결정할 수 없는 경우를 이러한 상황이라고 합니다. 함수 오버로딩 .
컴파일러가 모호성 오류를 표시하면 컴파일러는 프로그램을 실행하지 않습니다.
함수 오버로딩의 원인:
- 유형 변환.
- 기본 인수가 있는 함수입니다.
- 참조로 전달하는 함수입니다.
- 유형 변환:
간단한 예를 살펴보겠습니다.
#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<<> 어디에 반환 유형 함수가 반환하는 값의 유형입니다.
클래스_이름 수업의 이름입니다.
운영자 작전 op는 오버로드되는 연산자이고 연산자는 키워드인 연산자 함수입니다.
연산자 오버로딩 규칙
- 기존 연산자는 오버로드만 가능하고 새 연산자는 오버로드할 수 없습니다.
- 오버로드된 연산자에는 사용자 정의 데이터 유형의 피연산자가 하나 이상 포함되어 있습니다.
- 특정 연산자를 오버로드하기 위해 친구 함수를 사용할 수 없습니다. 그러나 멤버 함수를 사용하여 해당 연산자를 오버로드할 수 있습니다.
- 단항 연산자가 멤버 함수를 통해 오버로드되면 명시적인 인수를 취하지 않지만, 친구 함수에 의해 오버로드되면 하나의 인수를 취합니다.
- 이진 연산자가 멤버 함수를 통해 오버로드되면 하나의 명시적 인수를 취하고, friend 함수를 통해 오버로드되면 두 개의 명시적 인수를 사용합니다.
C++ 연산자 오버로딩 예
C++에서 연산자 오버로딩의 간단한 예를 살펴보겠습니다. 이 예에서는 void 연산자 ++() 연산자 함수가 정의됩니다(Test 클래스 내부).
// 단항 연산자 ++를 오버로드하는 프로그램입니다.
#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> 이항 연산자를 오버로드하는 간단한 예를 살펴보겠습니다.
// 이진 연산자를 오버로드하는 프로그램입니다.
#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>