Java 当Function.apply-Method'；它没有被覆盖吗？

在下面的示例代码中，我很难理解“apply”方法的用法。apply方法到底在做什么，因为它没有被覆盖？ 它显然返回了结果

ValidationResult

，但是没有代码支持它

public class Main {

    public static void main(String[] args) {
        
        Customer customer = new Customer("Alice", "alice@gmail.com", "0879546451", LocalDate.of(2000, 1, 1));
                
        ValidationResult result = CustomerRegistrationValidator
        .isEmailValid()
        .and(CustomerRegistrationValidator.isPhoneNumber())
        .and(CustomerRegistrationValidator.isAdult())
        .apply(customer);
        
        System.out.println(result);
    }
}

public interface CustomerRegistrationValidator extends Function<Customer, ValidationResult> {
    
    static CustomerRegistrationValidator isEmailValid() {
        return customer -> customer.getEmail().contains("@") ? ValidationResult.SUCCESS : ValidationResult.EMAIL_NOT_VALID;
    }
    
    static CustomerRegistrationValidator isPhoneNumber() {
        return customer -> customer.getPhoneNumber().contains("+08") ? ValidationResult.SUCCESS : ValidationResult.PHONE_NUMER_NOT_VALID;
    }
    
    static CustomerRegistrationValidator isAdult() {
        return customer -> Period.between(customer.getDob(), LocalDate.now()).getYears() > 16 ? ValidationResult.SUCCESS : ValidationResult.IS_NOT_AN_ADULT;
    }
    
    default CustomerRegistrationValidator and (CustomerRegistrationValidator other) {
        return customer -> {
            ValidationResult result = this.apply(customer);
            return result.equals(ValidationResult.SUCCESS) ? other.apply(customer) : result;
        };
    }
    
    enum ValidationResult {
        SUCCESS, PHONE_NUMER_NOT_VALID, EMAIL_NOT_VALID, IS_NOT_AN_ADULT;
    } 
}

公共类主{
公共静态void main（字符串[]args）{
客户=新客户（“Alice”alice@gmail.com“，“0879546451”，本地日期（2000年1月1日）；
ValidationResult=CustomerRegistrationValidator
.isEmailValid（）
.和（CustomerRegistrationValidator.isPhoneNumber（））
.和（CustomerRegistrationValidator.isadalt（））
.申请（客户）；
系统输出打印项次（结果）；
}
}
公共接口CustomerRegistrationValidator扩展函数{
静态CustomerRegistrationValidator isEmailValid（）{
return customer->customer.getEmail（）包含（“@”）？ValidationResult.SUCCESS:ValidationResult.EMAIL\u无效；
}
静态CustomerRegistrationValidator isPhoneNumber（）{
return customer->customer.getPhoneNumber（）包含（“+08”）？ValidationResult.SUCCESS:ValidationResult.PHONE\u NUMER\u无效；
}
静态CustomerRegistrationValidator isAdult（）{
return customer->Period.between（customer.getDob（），LocalDate.now（））.getYears（）>16？ValidationResult.SUCCESS:ValidationResult.IS不是成年人；
}
默认CustomerRegistrationValidator和（CustomerRegistrationValidator其他）{
退货客户->{
ValidationResult=this.apply（客户）；
返回result.equals（ValidationResult.SUCCESS）？other.apply（customer）：结果；
};
}
枚举验证结果{
成功，电话号码无效，电子邮件无效，不是成年人；
} 
}

apply方法到底在做什么，因为它没有被覆盖

取决于所选接口的静态方法。如果

isEmailValid

，则

apply

方法的主体为

customer.getEmail().contains("@") ? ValidationResult.SUCCESS : ValidationResult.EMAIL_NOT_VALID;

它显然返回了结果ValidationResult，但是没有代码支持它

public class Main {

    public static void main(String[] args) {
        
        Customer customer = new Customer("Alice", "alice@gmail.com", "0879546451", LocalDate.of(2000, 1, 1));
                
        ValidationResult result = CustomerRegistrationValidator
        .isEmailValid()
        .and(CustomerRegistrationValidator.isPhoneNumber())
        .and(CustomerRegistrationValidator.isAdult())
        .apply(customer);
        
        System.out.println(result);
    }
}

public interface CustomerRegistrationValidator extends Function<Customer, ValidationResult> {
    
    static CustomerRegistrationValidator isEmailValid() {
        return customer -> customer.getEmail().contains("@") ? ValidationResult.SUCCESS : ValidationResult.EMAIL_NOT_VALID;
    }
    
    static CustomerRegistrationValidator isPhoneNumber() {
        return customer -> customer.getPhoneNumber().contains("+08") ? ValidationResult.SUCCESS : ValidationResult.PHONE_NUMER_NOT_VALID;
    }
    
    static CustomerRegistrationValidator isAdult() {
        return customer -> Period.between(customer.getDob(), LocalDate.now()).getYears() > 16 ? ValidationResult.SUCCESS : ValidationResult.IS_NOT_AN_ADULT;
    }
    
    default CustomerRegistrationValidator and (CustomerRegistrationValidator other) {
        return customer -> {
            ValidationResult result = this.apply(customer);
            return result.equals(ValidationResult.SUCCESS) ? other.apply(customer) : result;
        };
    }
    
    enum ValidationResult {
        SUCCESS, PHONE_NUMER_NOT_VALID, EMAIL_NOT_VALID, IS_NOT_AN_ADULT;
    } 
}

在那里，一个

CustomerRegistrationValidator

实例由lambda定义

想象一下你必须重写这个

static CustomerRegistrationValidator isEmailValid() {
  return new CustomerRegistrationValidator() {
    @Override
    public ValidationResult apply(Customer customer) {
      return customer.getEmail().contains("@") ? ValidationResult.SUCCESS : ValidationResult.EMAIL_NOT_VALID;
    }
  };
}

---

到lambda语法。您将得到您现在所拥有的。

这是表示您希望将这3个函数（isEmailValid、isPhoneNumber、isAdult）应用于变量

customer

的语法。它展示了如何在一条语句中链接多个函数调用。