Question

我正在尝试让java函数接受一些输入，更改它并将其发回。通常你会返回varName;或者某些东西，但在这里它不起作用。我的问题是：如何创建一个自定义方法来接受变量'nameRaw'并在函数'nameChanger（）'中更改它然后将更改的名称放入变量：'nameChanged'。

我使用IDEONE.com显示代码并对其进行处理，所以这里是链接：http://ideone.com/cdj6Cd

如果您不信任随机链接，那就找到并且完全可以理解。所以，我也会把它放在这里，但只是假设用户输入的唯一内容是输入为“Test”。

代码：

/* package whatever; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
class Ideone
{
    public static void main (String[] args) throws java.lang.Exception {
        System.out.println("*INTENDED* Function of this-To create a custom method to take in a variable 'nameRaw' and change it within the function 'nameChanger()' then out put the changed name to variable: 'nameChanged'.\nProblem: the name changes within the function 'nameChanger()' only, therefore I cannot call the variable 'nameChanged' elsewhere, such as in main().........\n\n\n");
        // Initalize the Java Scanner
        Scanner in = new Scanner(System.in);

        // Initalizes a completely empty string with name "nameChanged"
        String nameChanged = null;

        System.out.println("Please enter a username for yourself below:");
        System.out.println("Test"); // Fake input to make me feel good I guess, it just looks better

        // Sets "nameRaw" to "Test"
        String nameRaw = in.nextLine();

        // Spits out the untouched value that the user has entered
        System.out.println("\nRaw name before passing to custom method: " + nameRaw);

        // Puts "nameRaw" into the custom method "nameChanger()" to change the, "s" in, "Test" to a "z" by default
        nameChanger(nameRaw, nameChanged);

        // Spits out the touched and changed nameChanged variable from "nameChanger()"
        if(nameChanged == null) {
            System.out.println("\nHere is the failed changed name: " + nameChanged);
            System.out.println("\nARE YOU KIDDING ME! WHY DOES THIS NOT WORK?!?! PLEASE HELP!!!!");
        } else {
            System.out.println("Here is the successfuly changed name: " + nameChanged);
            System.out.println("\nWhoever solved this problem is a god..");
        }

    } // Closes method main()

    // Custom method named "nameChanger" that will need a variable named "nameRaw" *which is set* within its () to function
    private static String nameChanger(String nameRaw, String nameChanged) {

        //// Initalizes a completely empty string with name "nameChanged"
        //String nameChanged = null;

        // States the set name *unchanged* from the main method on line 17
        System.out.println("\t#\tName unchanged read and displayed by the nameChanger custom method: " + nameRaw);

        // The name by default "Test" does contain an "s" so the if statement is satisfied
        if(nameRaw.contains("s")) {

            // The variable "nameRaw should be running through here and having it's "s" replaced with a "z"
            nameChanged = nameRaw.replace("s", "z");

            // Output the changed name *if all goes right*
            System.out.println("\t#\tName changed *still in custom function* is: " + nameChanged);

        // The else statement is just for testing purposes, such as changing "Test" as the input to "Demo" to dis-satisfy the if statemtent 
        } else {
            System.out.println("\t#\tFor reference, here is the unchanged name raw: " + nameRaw);
            System.out.println("\t#\tAlso for reference, here is the null value 'nameChanged': " + nameChanged);
        }

        // One more time to show my hate toward Java, output the changed variable "nameChanged". Take note that the "s" is a "z".....
        System.out.println("\t#\tPlease don't be null or have an 's' in it: " + nameChanged);

        // To output to main() that nameChanged is now not null *if all goes right*, but "Tezt" should be the output
        return nameChanged;
    } // Close custom method, nameChanger()
}

谢谢你们希望这不会让你感到太多：p，亚伦

Answer 1

简短回答：您需要在main方法中执行此操作：

nameChanged = nameChanger(nameRaw, nameChanged);

解释：如果更改对象引用在不同方法中指向的值，则调用方法将无法看到更改。在您的情况下，nameChanged方法中声明main并在nameChanger方法中重新分配，因此，main方法中无法看到更改。

但是，您可以通过调用它的可变方法来修改对象：

// in the main method
SomeStudent s = new SomeStudent();
modifyStudent(s);

// inside modifyStudent method
// change will be visible in the main method
s.setResult("PASS");

并且主要方法中会显示更改。

但是，在您的情况下，这是不可能的，因为Java中的String是不可变的。因此，您需要返回修改后的字符串并在main方法中捕获返回的值。

此外，由于上述原因，您无需将nameChanged传递到nameChanger。

Answer 2

由于Java方法调用始终是call-by-value，因此您无法通过在C中使用指针来更改当前块之外的变量（例如方法）。您需要使用返回值并分配它到变量

nameChanged = nameChanger(nameRaw, nameChanged);

或创建某种形式的holder对象，它在内部操作变量。

Answer 3

您正尝试通过副作用更改变量。请注意，当您调用nameChanger()时，实际上并未将返回值分配给变量。

String result = nameChanger(nameRaw)是一种更直接的方法。通常不希望依赖副作用。请注意，我还更改了方法签名，因为您现在不需要将其传递给您希望存储结果的变量，您可以在方法范围内创建它。

public static String nameChanger(String nameRaw) {
    String result = "";
    // do stuff
    if(nameRaw.contains("s")) {
        result = nameRaw.replace("s", "z");
    }
    //more stuff and System.out.println() statements
    return result;
}

Answer 4

如果我理解正确，你想创建一个方法，它将一个字符串作为参数，处理它然后返回更改后的值。

你离这儿不远：

/**
* NameChanger method takes a string as a parameter 
* and replaces its "s" with "z" then sends it back
*/
private static String nameChanger(String nameRaw) {
    String cReturnedString = nameRaw; 
    if(nameRaw.contains("s")) {
        cReturnedString = nameRaw.replace("s", "z");
    } 
    return cReturnedString;
}

// From another method
String cMyNewString = nameChanger("sissy");
System.out.println(cMyNewString); //Outputs "zizzy"

我改变了什么：

删除了额外的参数 - 您无需传递您在方法参数中返回的字符串。
初始化返回的字符串，其值与nameRaw变量相同。如果您的nameRaw不包含＆＃34; s＆＃34;，则您不希望返回null。

Answer 5

您不会传递可能会发生变化的变量。而是将代码更改为main()中的以下方法调用：

String nameChanged = nameChanger(nameRaw);

并在您实际创建它的方法中创建要返回的变量：

private static String nameChanger(String nameRaw) {
    String nameChanged = null; //
    // ...
    // and then
    return nameChanged;
}

Answer 6

这一行：

nameChanger(nameRaw, nameChanged);

需要像：

nameChanged = nameChanger(nameRaw, nameChanged);

之所以出现这种情况，是因为Java是一个值传递。所以你传递String的值而不是它的引用。

如果你真的，真的希望能够改变nameChanged; nameChanged需要是一个（通常是可变的）Object，它具有适当的setter / getter而不是native。在这种情况下，您将传递值'对象引用的副本'，它允许您在方法中更改对象的内容（如果它不是不可变的）

Answer 7

private static String nameChanger(String nameRaw, String nameChanged) 在这种方法中，你为什么要通过nameChanged？我认为nameChanged应该是nameChanger方法结束时返回的值。

我不明白为什么要检查nameChanged是否为空，但是您将其初始化为null并且没有更改它。

String nameChanged = null;


if(nameChanged == null) {
        System.out.println("\nHere is the failed changed name: " + nameChanged);
        System.out.println("\nARE YOU KIDDING ME! WHY DOES THIS NOT WORK?!?! PLEASE HELP!!!!");
    } else {
        System.out.println("Here is the successfuly changed name: " + nameChanged);
        System.out.println("\nWhoever solved this problem is a god..");
    }

如果你仍按照你的方法行事，请尝试 nameChanged = nameChanger(nameRaw, nameChanged) 而不是nameChanged = null

我希望这会有所帮助

Answer 8

谢谢大家的回答！你所说的一切都帮助我解决了这个问题！

我的解决方案：nameChanged = nameChanger(nameRaw, nameChanged)

感谢各位帮助，我非常感谢您的时间和脑力，向我解释这个以及为什么！

谢谢和最诚挚的问候，亚伦

让函数获取一些数据并将其发回

8 个答案: