String.valueOf（char）和+的区别

Question

to show the default value of char ,I wrote code like this:

public class TestChar {
  static char a;
  public static void main(String[] args) {
    System.out.println("."+String.valueOf(a)+".");
    System.out.println("the default char is "+a);
  }   
}

but the console output is confused.the first is ". ." ,however the second is "the default char is [](like this ,I don't know how to describe it.)" why?thanks for help

Answer 1

A char is not a String. In your first print statement you are converting the char to a string and then printing the value. In the second line you're just printing the value of the char directly.

Answer 2

As user1681360 mentioned, it is the character '\\0' you are printing. You have not initialized your field, so Java initializes it for you to the default value, unprintable character '\\0' . Depending on the environment, unprintable characters will be shown as empty boxes, question marks, etc.

In the first line you are first creating a String , then appending it. In the second line the operator String + char is compiled to java.lang.StringBuilder#append(char) , so you are directly appending a character to the buffer, rather that creating a temporary String .

Indeed the two approaches are always equivalent, even when the character in question is '\\0' . The character '\\0' has a special treatment in Java Language Specification, but that does not affect this particular behavior.

Answer 3

You haven't initialized the value of a, so it is equivalent to char a = 0; a non-printable character.

Initialize the value to something lets say static char a = 'a'; That should show you something.

Update: as the second character is non-printable it is displayed as a square but it is in fact a '\\0' character.

Update 2: The default value is 0, as in the number 0, not the character '0'.