Displaying Char And Integer Together

Question

My code is in Java. I have a char array that has int values as well as char values. When I print, I get its values related to ASCII. But when I convert output to int, my character value changes to its ASCII value.

char arr[]=new char [5];
for(int i=0;i<5;i++)
{
    arr[i]=(char)(i+1);
}

arr[0]='@';

for(int i=0;i<5;i++)                
{
    System.out.println(arr[i]);
}

Or converting to int:

for(int i=0;i<5;i++)                
{
   System.out.println((int)arr[i]);
}

My output is

64 2 3 4 5

or

@

depending on which code I use. However, I want the output as

@ 2 3 4 5

I tried using method overloading to separate integers and characters, but that didn't work.

Is there a condition for int like below?

if(arr[i]==int)

The code doesn't work but is there some thing similar to it or any other solution?

Answer 1

It doesn't work, because as far a Java is concerned, a char primitive is just a 16 bit unsigned integer value. To demonstrate:

System.out.println('A' == 65); // prints true

One way to work around this would be to declare an object array, and rely on auto-boxing to have Java automatically convert your primitive value to an Integer or Character wrapper object:

Object arr[] = new Object[2];
arr[0] = '@'; // stores '@' as a Character wrapper object
arr[1] = 1; // stores 1 as an Integer wrapper object

for (Object o : arr) {
    System.out.println(o);
}

Unlike primitives, wrapper objects are aware of their type, so this prints out:

@ // by calling Character.toString()
1 // by calling Integer.toString()

Note that you lose some compile-time checking in the process. Your object array will not only accept Character and Integer values, but also any other value.