Why does the toString method in java not seem to w

问题:

I want to convert a character array to a string object using the toString() method in java. Here is a snippet of the test code I used:

import java.util.Arrays;
class toString{
    public static void main(String[] args){
        char[] Array = {\'a\', \'b\', \'c\', \'d\', \'e\', \'f\'};
        System.out.println(Array.toString());
        }
}

In principle, it should print abcdef, but it is printing random gibberish of the likes of [C@6e1408 or [C@e53108 each time the program executes. I don\'t need an alternative out of this but want to know why this is happening.

回答1:

To get a human-readable toString(), you must use Arrays.toString(), like this:

System.out.println(Arrays.toString(Array));

Java\'s toString() for an array is to print [, followed by a character representing the type of the array\'s elements (in your case C for char), followed by @ then the \"identity hash code\" of the array (think of it like you would a \"memory address\").

This sad state of affairs is generally considered as a \"mistake\" with java.

See this answer for a list of other \"mistakes\".

回答2:

I don\'t know where you get the idea that \"in principle\" it should print \"abcdef\". Where is that documented?

Something like [C@6e1408 is certainly not random gibberish - it\'s the same way of constructing a string from an object as any other type that doesn\'t override toString() inherits - it\'s a representation of the type ([ indicating an array; C indicating the char primitive type) followed by the identity hash code in hex. See the documentation for Object.toString() for details. As it happens, arrays don\'t override toString.

If you want [a, b, c, d, e, f] you can use Arrays.toString(char[]). If you want abcdef you can use new String(char[]).

回答3:

Just use the following commands to get your abcdef array printed

    String a= new String(Array);

    System.out.println(a);

there you have it problem solved !! now regarding why is printing the other stuff i think those guys above put some useful links for that. Ok gotta go !!

回答4:

Because a char array is an array of primitives and toString() will give you it\'s default (which is a hash of the object). Some classes will implement toString() to do cooler things, but primitaves will not.

回答5:

Arrays don\'t override toString. There\'s a static method: java.util.Arrays.toString that should solve your problem.

import java.util.Arrays;
class toString {
    public static void main(String[] args){
        char[] Array = {\'a\', \'b\', \'c\', \'d\', \'e\', \'f\'};
        System.out.println(Arrays.toString(Array));
    }
}

回答6:

The default implementation of the toString method of the char [] class returns a String representation of the base address of the array, which is what is being printed here. We cannot change it, since the class of char [] is not extendable.

回答7:

char[] Array = { \'a\', \'b\', \'c\', \'d\', \'e\', \'f\' };
System.out.println(Array);

It should print abcdef.

回答8:

There is a spelling mistake of \"Array.toString()\" to \"Arrays.toString(Array)\" I guess so, and instead of writing name.toString(), pass the name as an argument and Write as above.

回答9:

this way I found worked:

public String convertToString(char[] array, int length)
{
String char_string;
String return_string=\"\";
int i;

   for(i=0;i<length;i++)
   {
       char_string=Character.toString(array[i]);
       return_string=return_string.concat(char_string);
   }
   return return_string;
}