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HomeJavajavaTutorialWhat is the handling of different parameter types in the Java function overloading mechanism?

Rules for handling different parameter types in Java function overloading: Exact match: This method is used when there is an overloaded method whose parameter list exactly matches the actual parameter type. Widening conversion: When there is no exact match, try to convert the actual parameter type to a wider type. Boxing/Unboxing: Automatic boxing or unboxing between primitive types and wrapped classes. Variable parameters: Variable parameters (...) can match any number of parameters of the same type.

Java 函数重载机制中不同参数类型的处理方法是什么?

Different parameter type processing mechanism in Java function overloading

Function overloading is a method in Java that allows the creation of functions with the same Ability to have multiple methods with different names but different parameter lists. When an overloaded method is called, the Java compiler determines the specific method to call based on the actual parameter types provided in the call.

The overloading rules for function overloading in Java are as follows:

  • The method names must be the same.
  • Method parameter lists must be different, either in number, type or order.
  • The return value types can be the same or different.

Handling of different parameter types

When processing overloaded methods of different parameter types, the Java compiler matches according to the following rules:

  • Exact match: If an overloaded method's parameter list is found to exactly match the actual parameter types supplied in the call, the compiler will select that method.
  • Wide conversion: If an exact match is not found, the compiler will try to convert the actual parameter type to a wider type (such as converting int to long).
  • Autoboxing/unboxing: The Java compiler automatically performs boxing and unboxing between primitive types and their corresponding wrapper classes.
  • Variable parameters: Variable parameters (...) in Java can match any number of parameters of the same type.

Practical case

Consider the following example class in which the add method is overloaded multiple times:

class Calculator {

    public int add(int a, int b) {
        return a + b;
    }

    public double add(double a, double b) {
        return a + b;
    }

    public long add(long a, long b) {
        return a + b;
    }

}

Example of calls:

Calculator calculator = new Calculator();

int result1 = calculator.add(10, 20); // 调用 int 参数的 add() 方法
double result2 = calculator.add(10.5, 15.3); // 调用 double 参数的 add() 方法
long result3 = calculator.add(1000L, 2000L); // 调用 long 参数的 add() 方法

In these calls, the compiler chooses the correct overloaded method based on the supplied argument types:

  • result1 Call the add method with an int parameter because the actual parameter type is int.
  • result2 Calls the add method of the double parameter because the actual parameter type is double.
  • result3 Calls the add method for the long parameter because the actual parameter type is long.

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