在另一个void函数内部传递一个私有void函数（编译错误Java）

Question

Let's say I have a function which makes the names into first letter uppercase and the rest in lowercase when a parameter is entered.

I have another function to system print other things including the function I mentioned above. But I keep on getting compile error saying 'void' type not allowed here.

Still really new with Java, not sure how I can adjust it. Can someone give me a hand?

This is the first function I mentioned

private void makePrettyString(String modelName)
    {

        // Change first letter of the name into upper case and store the first char only
        char first = Character.toUpperCase(modelName.charAt(0));

        // lower case the whole name
        String lower = modelName.toLowerCase();

        // store the whole name except the first char into the variable
        String restInLowerCase = lower.substring(1);

        // Combine the first char (which is upper case) and the rest of the name (which is in lower case)
        this.modelName = first+restInLowerCase; 
    }

The second function to get some details

public void getCarDetails()
{
    System.out.println(makePrettyString(modelName));
}

Answer 1

Thank you all for the help! I just realized this is all I need to do change the method from void to String and of course as everyone says return something So this works

private String makePrettyString(String modelName)
{

    // Change first letter of the name into upper case and store the first char only
    char first = Character.toUpperCase(modelName.charAt(0));

    // lower case the whole name
    String lower = modelName.toLowerCase();

    // store the whole name except the first char into the variable
    String restInLowerCase = lower.substring(1);

    // Combine the first char (which is upper case) and the rest of the name (which is in lower case)
    return = first+restInLowerCase; 
}

Answer 2

Modify the function as:

private String makePrettyString(String modelName) {

   //Your existing code

   return this.modelName;
}

With the above your job gets resolved as the udner the System.out.println(fun-call) the argument is expecting some value as written which unfortunately is nothing but void!!

Answer 3

change void of first method to string. it has to return a string so change the method signature to string

like this :

private String makePrettyString(String modelName)     // <-- change has to be made here  (String instead of void)
{

    // Change first letter of the name into upper case and store the first char only
    char first = Character.toUpperCase(modelName.charAt(0));

    // lower case the whole name
    String lower = modelName.toLowerCase();

    // store the whole name except the first char into the variable
    String restInLowerCase = lower.substring(1);

    // Combine the first char (which is upper case) and the rest of the name (which is in lower case)
    return = first+restInLowerCase; 
}

Answer 4

The line --> System.out.println(makePrettyString(modelName)); will cause the error because the class PrintStream in which println() is defined doesn't have an overloaded method of println() which takes void as argument.

Check all overloaded methods here .

Just to clear the mist note that you cannot have a method that accepts void as argument in Java. Now, there is a place holder / wrapper for void which goes by the name Void which *also needs an argument of type Void .

Answer 5

If both methods getCarDetails() and makePrettyString() are in the same class,

Simply write
makePrettyString(); System.out.println(this.modelName) makePrettyString(); System.out.println(this.modelName) ;

Instead of using a parameter makePrettyString(String modelName) , you can access this.modelName in the makePrettyString() method.

HTH.