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How to reconvert from int to AScII in java? [duplicate]

This question already has answers here:
Convert int to char in java
(18 answers)
Closed 1 year ago.
So what i understand is that if for example i have an int a = 49 and want to find out the ASCII equivalent to it, all i need to do is:
int a = 49;
System.out.println((char)a);
which has the Output: 1.
But how can i do this reversed? So lets say i have int a = 1 and i want the output to be 49?
I have already tried stuff like:
int a = 1;
System.out.println ((char)a);
But the output here is "." instead of 49.

This:
int a = 49;
System.out.println((char)a);
gives you the char whose internal value is 49, which is to say the unicode character '1'. There is no numerical conversion going on, it's just two ways to look at the same number.
In this:
int a = 1;
System.out.println((char)a);
you get the char whose internal value is 1, which is a unicode control character (ctrl/A, which probably won't be printed in any useful way). There is nothing there that can magically come up with the value 49.
If you had the character '1' then it's easy:
int a = '1';
System.out.println(a);
but this is practically the same as your first case; '1' and 49 are the same value.

Why is java allowing this to compile and what is it interpreting multiple + signs as in this case? [duplicate]

This question already has answers here:
Why does this Java code with "+ +" compile?
(8 answers)
Explanation about a Java statement
(6 answers)
Closed 4 years ago.
Wasn't sure exactly how to word the question, but I noticed something strange while constructing a date. I found that if I construct a date like this
new Date(+ 1)
it compiled just fine, and so did
new Date(+ + + 1)
If I execute the following, the output is 1
public static void main(String[] args) {
int x = 1;
System.out.println(+ + + + x);
}
Can anyone explain what it is that the JVM thinks I am doing?

It's the unary operator (+). You can always add a + to a numeral and that will give you the positive value of the number.
Because you're spacing the tokens out in such a fashion, the lexer is not interpreting anything here as incrementation, so you're adding four unary (+) operations to a value 1.

It's treating it like this:
System.out.println(+ (+ (+ (+ x))));
This is no different than
System.out.println(- (- (- (- x))));

Adding three different char in java as java variable and getting a number [duplicate]

This question already has answers here:
In Java, is the result of the addition of two chars an int or a char?
(8 answers)
Closed 4 years ago.
IntelliJ IDEA Capture
Why i am getting 152, I think it will give me an error.
Please explain it.
public class character {
public static void main(String[] args) {
char myCharValue1 = 'A';
char myCharValue2 = '2';
char myCharValue3 = '%';
System.out.println(myCharValue1 + myCharValue2 + myCharValue3);
}
}

That is because chars refer to a number, which in turn has an ASCII representation.
Looking at an ASCII table you can see that the chars A, 2 and % have following values respectivly: 65, 50 and 37.
Adding those numbers together, you'll end up with 152 which is what you got in your example.
To print out those chars you could use following:
System.out.printf("%s%s%s&n", myCharValue1 + myCharValue2 + myCharValue3);
Which will print A2% (and a newline)

The concatenation + is for String. What you're doing is adding the numeric values of your chars and printing them together.
If you start with "" and then use + as Patrick Parker shows in his comment, it will become concatenation instead of simple addition and you'll get the result you expect.

Why does a char disappear when concatinated with an integer [duplicate]

This question already has answers here:
The concatenation of chars to form a string gives different results
(5 answers)
why does a char + another char = a weird number
(4 answers)
Closed 8 years ago.
We had an odd thing in our logging happen today. Printed is a list of integers and longs separated by commas, like the following code:
public class Main {
public static void main(String[] args) throws InterruptedException {
long l = 10;
System.out.println(l + ';' + "text");
}
}
The problem was that the ; disappeared from the output.

The problem here is caused by the overload of the + operator. It acts in one way when operating on a String and a long, and another when operating on a char and a long. When one of the operands is a string it will try to cast the other operand to a string if it's not already one and then concatenate the two.
But when the operators are numbers, like int and long, the + operator acts as the normal mathematical plus operator. And since char is a number and nothing else, ';' + l is treated as a numerical operation and thus the output of the code in the question is 69text and not 10;text.

Creating a random 4 digit number, and storing it to a string [duplicate]

This question already has answers here:
How to Convert an int to a String? [duplicate]
(6 answers)
Closed 8 years ago.
I'm trying to create a method which generates a 4 digit integer and stores it in a string.
The 4 digit integer must lie between 1000 and below 10000. Then the value must be stored to PINString.
Heres what I have so far. I get the error Cannot invoke toString(String) on the primitive type int. How can I fix it?
public void generatePIN()
{
//generate a 4 digit integer 1000 <10000
int randomPIN = (int)(Math.random()*9000)+1000;
//Store integer in a string
randomPIN.toString(PINString);
}

You want to use PINString = String.valueOf(randomPIN);

Make a String variable, concat the generated int value in it:
int randomPIN = (int)(Math.random()*9000)+1000;
String val = ""+randomPIN;
OR even more simple
String val = ""+((int)(Math.random()*9000)+1000);
Can't get any more simple than this ;)

randomPIN is a primitive datatype.
If you want to store the integer value in a String, use String.valueOf:
String pin = String.valueOf(randomPIN);

Use a string to store the value:
String PINString= String.valueOf(randomPIN);

Try this approach. The x is just the first digit. It is from 1 to 9.
Then you append it to another number which has at most 3 digits.
public String generatePIN()
{
int x = (int)(Math.random() * 9);
x = x + 1;
String randomPIN = (x + "") + ( ((int)(Math.random()*1000)) + "" );
return randomPIN;
}