Variable type conversion

Question

I couldn't assign int variables into char[] array

I tried type casting, but it didn't work as I expected
the complier keeps return me ASCII value which I didn't want them. I want the numbers the actual number

char[] char_array = new char[10];
for(int i = 0; i < 10; i ++)
{
  int calculation = i * (d - 1);
  char_array[i] = (char) calculation;
}

After print out
It gives me something like this

| B @ #

If you want the numbers, store them into an `int[]`. You're using `char[]`, which contains characters, and by default they're printed as characters instead of their numeric value. — Kayaman, Sep 09 '19 at 11:08
Using System.out.print(calculation) returns aal the numbers just fine, but the problem is those values coudln't be converted into char type. Are there any ways to fix this? Thhank you. — user3505908, Sep 09 '19 at 11:09
No, i *8 /2 ...; is just the example. it was actually d * (d - 1) — user3505908, Sep 09 '19 at 11:10
What do you mean by "something like this `| B @ #`"? Is it something like `[B@ab47d2`? If so take a look at [What's the simplest way to print a Java array?](https://stackoverflow.com/q/409784) — Pshemo, Sep 09 '19 at 11:53
If that is not the case then please provide [mcve] (a.k.a. [SSCCE](http://sscce.org)) and *actual* results you are getting. Also include what *other* result you ware expecting (and why). Without it we need to play guessing game which is not very efficient way of solving problems. — Pshemo, Sep 09 '19 at 11:59

score 2 · Answer 1 · 2019-09-10T08:23:28.773

You can't use char to store numeric values. Its because when your store an int value to a char variable type conversion occurs as follows:

char i = 60 + 5;

Here, i will store A which has Unicode value 65.

Similarly, when you store a char value to an int, type conversion occurs as follows:

int i = 'A'

This will store 65;

This is how java performs type cast between int and char;

So, to deal with this situation, use the following code:

int[] int_array = new int[10];
for(int i = 0; i < 10; i ++)
{
int calculation = i *8 /2 ...;
int_array[i] = calculation;
}

or you can also use double if you want precise results;

Java doesn't use ASCII. `char` is a UTF-16 code unit. UTF-16 is one of several character encodings for the Unicode character set. — Tom Blodget, Sep 09 '19 at 22:33

score 0 · Answer 2 · answered Sep 09 '19 at 11:09

0

try, this should help you.

int a = 1
char b = Integer.toString(a).charAt(0);
System.out.println(b);

answered Sep 09 '19 at 11:09

Jayant Varshney

1,765
1
25
42

I just realized that char only accept 1 position value. – user3505908 Sep 09 '19 at 11:12

score 0 · Answer 3 · answered Sep 09 '19 at 11:11

For printing number, you should cast the output to int.

char[] char_array = new char[10];
for(int i = 0; i < 10; i ++)
{
  int calculation = i * 4;
  char_array[i] = (char) calculation;
}

for(int i = 0; i < 10; i ++)
{
    System.out.println( (int) char_array[i] );
}

But in this case consider to use a int[] directly instead casting forth and back.

Also note, that overflow can happen when storing char c = (int) 256.

Variable type conversion

3 Answers3