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Adding string literals to initialized variables

I feel silly that I can't find the answer to this question, but I'm working on an assignment for class and I'm asked to describe the output of the following sample code:
int i = 1;
for (i = 0; i <= 6; i++){
System.out.print( 'i' + i);
}
which outputs:
105106107108109110111
((I understand that initializing i to 1 is not necessary before the loop condition))
I don't understand why the above print statement outputs this pattern of numbers (1 05 1 06 1 07 1 08 1 09 1 10 1 11). Simply leaving it as
System.out.print( 'i');
prints "i" 5 times as expected. So why does adding the value of i change the output of 'i'?
edit: fixed variable name

Because 'i' is a character literal of type char. Adding a char value and an int value automatically promotes it to an int. The ASCII value of the lower case i is 105 (0x69 in hex).
So what you have is System.out.print(105+i) etc.
If you want to concatenate strings, you have to use strings: System.out.print("i" + i) or System.out.printf("i%d", i). If your char were dynamic and stored in variable with name c, you might want to use String.valueOf(c) + i or printf/String.format again.

j appears to be uninitialised in this snippet, and so it shouldn't even compile. If you have a j variable somewhere else in the scope your for loop will be using that.
Also note that 'j' + j is trying to add a (presumably) integer to a char, which will promote it to an int, and so you are printing the integer code point of 'j' plus whatever variable j is.

in java you can add char with int and the character implicitly will convert to corresponding ASCII code.
in your print statement you are using single quotes, so it will be infer as character.
also notice the ASCII code for alphabet i is equal to 105.

Java syntax int from char magic [duplicate]

This question already has answers here:
Java: Subtract '0' from char to get an int... why does this work?
(10 answers)
How does subtracting the character '0' from a char change it into an int?
(4 answers)
Closed 8 years ago.
I’m learning Java through "introduction to Java programming 9th edition" by Daniel Liang at chapter 9 "strings" I’ve encountered this piece of code :
public static int hexCharToDecimal(char ch) {
if (ch >= 'A' && ch <= 'F')
return 10 + ch - 'A';
else
return ch - '0';
}
Can someone explain what just happened in here? How is possible to add/subtract chars from integers and what's the meaning behind it?

From the Docs
The char data type is a single 16-bit Unicode character.
A char is represented by its code point value:
min '\u0000' (or 0)
max: '\uffff' (or 65,535)
You can see all of the English alphabetic code points on an ASCII table.
Note that 0 == \u0000 and 65,535 == \uffff, as well as everything in between. They are corresponding values.
A char is actually just stored as a number (its code point value). We have syntax to represent characters like char c = 'A';, but it's equivalent to char c = 65; and 'A' == 65 is true.
So in your code, the chars are being represented by their decimal values to do arithmetic (whole numbers from 0 to 65,535).
For example, the char 'A' is represented by its code point 65 (decimal value in ASCII table):
System.out.print('A'); // prints A
System.out.print((int)('A')); // prints 65 because you casted it to an int
As a note, a short is a 16-bit signed integer, so even though a char is also 16-bits, the maximum integer value of a char (65,535) exceeds the maximum integer value of a short (32,767). Therefore, a cast to (short) from a char cannot always work. And the minimum integer value of a char is 0, whereas the minimum integer value of a short is -32,768.
For your code, let's say that the char was 'D'. Note that 'D' == 68 since its code point is 68.
return 10 + ch - 'A';
This returns 10 + 68 - 65, so it will return 13.
Now let's say the char was 'Q' == 81.
if (ch >= 'A' && ch <= 'F')
This is false since 'Q' > 'F' (81 > 70), so it would go into the else block and execute:
return ch - '0';
This returns 81 - 48 so it will return 33.
Your function returns an int type, but if it were to instead return a char or have the int casted to a char afterward, then the value 33 returned would represent the '!' character, since 33 is its code point value. Look up the character in ASCII table or Unicode table to verify that '!' == 33 (compare decimal values).

This is because char is a primitive type which can be used as a numerical value. Every character in a string is encoded as a specific number (not entirely true in all cases, but good enough for a basic understanding of the matter) and Java allows you to use chars in such a way.
It probably allows this mostly for historical reasons, this is how it worked in C and they probably motivated it with "performance" or something like that.
If you think it's weird then don't worry, I think so too
The other answer is incorrect actually. ASCII is a specific encoding (an encoding is some specification that says "1 = A, 2 = B, ... , 255 = Space") and that is not the one used in Java. A Java char is two bytes wide and is interpreted through the unicode character encoding.

Chars are in turn stored as integers(ASCII value) so that you can perform add and sub on integers which will return ASCII value of a char

Regardless of how Java actually stores the char datatype, what's certain is this, the character 'A' subtracted from the character 'A' would be represented as the null character, \0. In memory, this means every bit is 0. The size in memory a char takes up in memory may vary from language to language, but as far as I know, the null character is the same in all the languages, every bit is equal to 0.
As an int value, a piece of memory with every bit equal to 0 represents the integer value of 0.
And as it turns out, when you do "character math", subtracting any alphabetical character from any other alphabetical character (of the same case) results in bits being flipped in such a way that, if you were to interpret them as an int, would represent the distance between these characters. Additionally, subtracting the char '0' from any other numeric char will result in int value of the char you subtracted from, for basically the same reason.
'A' - 'A' = '\0'
'a' - 'a' = '\0'
'0' - '0' = '\0'

How to convert char 'A' to binary '1010' in java language

I want to convert character type to binary but not char byte to bit.
I'll show you example.
if char = 'A' but actually it's hex! A = 1010
I have a character type but I want to represent it like hexadecimal.
Of course, I have only character data that can be matched hexadecimal.
like 9, 8, A, C, D not 47, U, Y ..
And Can I count size or length of 1 in 1010 ?
If you know how to solve this please let me know thank you!

Simple Steps
Here you can convert any Alphabet to the binary number
System.out.println(Long.toString(Long.parseLong("A", 16), 2)); //Output 1010
Thanks

You can most easily (and probably most efficiently) use Character.digit().
int a = Character.digit('A', 16);
System.out.println(Integer.toBinaryString(a));

Since the input is a char, convert using the helper method from Character:
int num = Character.getNumericValue(ch);
Then convert it to a binary string:
String s = Integer.toBinaryString(num);

Output of System.out.println( m[1]+` ´+m[ 2]) [duplicate]

To my understanding a char is a single character, that is a letter, a digit, a punctuation mark, a tab, a space or something similar. And therefore when I do:
char c = '1';
System.out.println(c);
The output 1 was exactly what I expected. So why is it that when I do this:
int a = 1;
char c = '1';
int ans = a + c;
System.out.println(ans);
I end up with the output 50?

You're getting that because it's adding the ASCII value of the char. You must convert it to an int first.

Number 1 is ASCII code 49. The compiler is doing the only sensible thing it can do with your request, and typecasting to int.

You end up with out of 50 because you have told Java to treat the result of the addition as an int in the following line:
int ans = a + c;
Instead of int you declare ans as a char.
Like so:
final int a = 1;
final char c = '1';
final char ans = (char) (a + c);
System.out.println(ans);

Because you are adding the value of c (1) to the unicode value of 'a', which is 49. The first 128 unicode point values are identical to ASCII, you can find those here:
http://www.asciitable.com/
Notice Chr '1' is Dec 49. The rest of the unicode points are here:
http://www.utf8-chartable.de/

A char is a disguised int. A char represents a character by coding it into an int. So for example 'c' is coded with 49. When you add them together, you get an int which is the sum of the code of the char and the value of the int.

'1' is a digit, not a number, and is encoded in ASCII to be of value 49.
Chars in Java can be promoted to int, so if you ask to add an int like 1 to a char like '1', alias 49, the more narrow type char is promoted to int, getting 49, + 1 => 50.
Note that every non-digit char can be added the same way:
'a' + 0 = 97
'A' + 0 = 65
' ' + 0 = 32

'char' is really just a two-byte unsigned integer.
The value '1' and 1 are very different. '1' is encoded as the two-byte value 49.
"Character encoding" is the topic you want to research. Or from the Java language spec: http://docs.oracle.com/javase/specs/jls/se7/html/jls-4.html#jls-4.2.1

char to int conversion

So I have something like this:
char cr = "9783815820865".charAt(0);
System.out.println(cr); //prints out 9
If I do this:
int cr = "9783815820865".charAt(0);
System.out.println(cr); //prints out 57
I understand that the conversion between char and int is not simply from '9' to 9. My problem is right now I simply need to keep the 9 as the int value, not 57. How to get the value 9 instead of 57 as a int type?

You can try with:
int cr = "9783815820865".charAt(0) - '0';
charAt(0) will return '9' (as a char), which is a numeric type. From this value we'll just subtract the value of '0', which is again numeric and is exactly nine entries behind the entry of the '9' character in the ASCII table.
So, behind the scenes, the the subtraction will work with the ASCII codes of '9' and '0', which means that 57 - 48 will be calculated.

try this:
char c = "9783815820865".charAt(0);
int cr = Integer.parseInt(c+"");

Using Character#getNumericValue may be more idiomatic. Bear in mind that it'll convert anything above 'A' as 10.
int cr = Character.getNumericValue("9783815820865".charAt(0));
System.out.println(cr);