Question

使用代码比使用单词更容易理解这个问题：

Map<Integer, Parent> objectMap = new HashMap<Integer, Parent>();

Parent myParent;
Child1 myChild1;
Child2 myChild2;
//A lot more myChilds

objectMap.put(1, myChild1);
objectMap.put(2, myChild2);
//Place all the myChilds

myChild1 = new Child1();  //Constructor is expensive, object may not get used
myChild2 = new Child2();  //Constructor is expensive, object may not get used
//Call constructor for all of myChilds

Parent finalObject;

int number = 1; //This can be any number

finalObject = objectMap.get(number);

如您所见，我事先并不知道finalObject会是哪一个类。代码工作没有问题，但这是我的问题：

如何避免同时调用两个构造函数？

由于只使用myChild1或myChild2且构造函数方法非常昂贵，我只想调用实际使用的那个。

像

这样的东西

finalObject.callConstructor();

在最后一行

有什么想法吗？

提前致谢。

编辑：我想知道的是如何在不知道类的名称的情况下调用构造函数。检查更新的代码。

Answer 1

这个怎么样？

#include <stdio.h>
#include <stdlib.h>
char getStaticElement();
char getDynamicElement();

int main() {
    char dynamicElement = getDynamicElement();
    char staticElement = getStaticElement();
    printf("Dynamic Element: %c\n", dynamicElement);
    printf("Static Element: %c\n", staticElement);
    return 0;

}
char getStaticElement() {
    char staticArray [] = {'a','b','c'};
    return staticArray[1]; // returns b
}
char getDynamicElement() {
    char * dynamicArray = malloc(sizeof(char)*4);
    dynamicArray [0] ='a';
    dynamicArray [1] ='b';
    dynamicArray [2] ='c';
    dynamicArray [3] ='\0';
    return dynamicArray[1]; // returns b
}

或者，甚至更好，这个？

Parent finalObject;
if (condition) {
    finalObject = new Child1();
} else {
    finalObject = new Child2();
}

Answer 2

不要构建两个对象。只构造你需要的对象。

<div id='donor_table'>

    <div style='height: 600px; width: 100%; overflow: scroll'>

        <table id='donor_tbl' border='5' cellpadding='10'>
            <thead>
            <th>#</th>
            <th>Announcements</th>
            <th>Date</th>
            <th>Time</th>
            <th>Post</th>
            </thead>
            <tbody>

            <?php
            include "connection.php";
            error_reporting(0);
            $searchannouncement = $_POST['searchannouncement'];
            $displayAnn = " SELECT annid,announcement,dateee,timeee FROM announcements_tbl WHERE annid LIKE '%" . $searchdonor . "%' OR announcement LIKE '%" . $searchdonor . "%' OR dateee LIKE '%" . $searchdonor . "%' OR timeee LIKE '%" . $searchdonor . "%' order by annid desc";

            $annData = mysql_query($displayAnn);
            while ($drecords = mysql_fetch_assoc($annData)) {
                echo "<tr><td>" . $drecords['annid'] . "</td>";
                echo "<td>" . $drecords['announcement'] . "</td>";
                echo "<td>" . $drecords['dateee'] . "</td>";
                echo "<td>" . $drecords['timeee'] . "</td>";
                echo "<td align='center'><input type='button' class='btn-post-announcement button' id=" . $drecords['annid'] . " value='Post'></td>";
            };
            ?>
            </tbody>
        </table>
    </div>
</div>
</body>
</html>
<div id="dlg">
    <div>
        <script type="text/javascript">
            $(function () {
                $("#dlg").dialog({
                    autoOpen: false,
                    height: 600,
                    width: 780,
                    modal: true
                });
                $(".btn-post-announcement").button();
                $(".btn-post-announcement").click(function () {
                })
                $("#donor_tbl").DataTable();
            });
        </script>

Answer 3

您应该使用工厂模式：

public interface Factory<T> {
  T create();
}

...

Factory<T> factory;
if (condition) {
  factory = FACTORY1;
} else {
  factory = FACTORY2;
}
object = factory.create()

Answer 4

首先检查你的条件，然后在if或else语句中调用构造函数。如果（条件） finalobject = new myChild1（）; 其他 finalobject = new myChild2（）;

在不知道类名的情况下调用构造函数（java）

4 个答案: