我试图让一个java函数接受一些输入，更改它并将其发送回来。通常你会 return varName;或其他东西，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，然后将更改后的名称输出到变量：“nameChanged”。
我已经使用IDEONE.com来展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，这是完全可以理解的。因此，我也将其放在这里，但假设用户输入的唯一内容是“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()
}
谢谢大家，希望这不会让你们不知所措:p, 亚伦
您不传递可能发生变化的变量。相反，请将代码更改为
main()String nameChanged = nameChanger(nameRaw);
并在实际创建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;
}
这一行：
nameChanger(nameRaw, nameChanged);
需要类似于：
nameChanged = nameChanger(nameRaw, nameChanged);
之所以会出现这种情况，是因为Java是按值传递的。因此，您将传入字符串的值而不是其引用。
如果你真的、真的希望能够改变
nameChangednameChanged简短回答： 您需要在您的
mainnameChanged = nameChanger(nameRaw, nameChanged);说明：如果在不同的方法中更改对象引用指向的值，则调用方法将看不到更改。在您的情况下，
nameChangedmainnameChangermain但是，您可以通过调用对象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");
并且更改将在 main 方法中可见。
但是，在您的情况下这是不可能的，因为 Java 中的
Stringmain此外，由于上述原因，您不需要将
nameChangednameChangerprivate static String nameChanger(String nameRaw, String nameChanged)nameChangednameChangednameChanger我不明白为什么你要检查
nameChangednullString 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..");
}
如果您仍然按照您的方法，请尝试
nameChanged = nameChanger(nameRaw, nameChanged)nameChanged = null我希望这有帮助
由于在 Java 中方法调用始终是按值调用，因此您无法像在 C 中使用指针那样更改当前块（例如方法）之外的变量。您需要使用返回值并分配它到变量
nameChanged = nameChanger(nameRaw, nameChanged);
或创建某种形式的持有者对象，在内部操作变量。
您试图通过副作用来更改变量。请注意，当您调用
nameChanger()String result = nameChanger(nameRaw)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;
}
谢谢所有回答的人！ 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)谢谢你们的帮助，我非常感谢你们花时间和脑力向我解释这一点以及原因！
谢谢并致以最诚挚的问候， 亚伦