Int to String

[Wolfgang Obi]

quick question....

is it possible for me in any way to read in a natural integer consisting of say 2 or more digits and then "cast" this into a string ? (i know this sounds stupid but i cant find better words with which to express my thoughts )

or

better still: convert it into a string or int array (such that each digit counts as an array component)....

...so i can index each digit....and apply methods such as length(); etc etc....

I'd appreciate a helpin hand.

thx

W.O.

[Edwin Dalorzo]

How about this?

int myNumber = 28;

//into String
String myString = String.valueOf(myNumber);

//into an char array
char[] myChars = myString.toCharArray();

[Wolfgang Obi]

thank you very much

i've decided to use String.valueOf();

PS: is it also possible to make use of the toString(); method in this context ?

[Robert Hill]

String myString = ""+myNumber;

[Edwin Dalorzo]

Well you could do this:

 
int myInt = 28;
String myString = Integer.valueOf(myInt).toString();
 

...or if you are using Tiger (JDK 1.5), you can take adventage of autoboxing. Like this:

Integer myAge = 28; //Autoboxing
myAge.toString();

[ May 06, 2006: Message edited by: Edwin Dalorzo ]

[Wolfgang Obi]

@Robert Hill: Thanx very simple and uncomplicated...., what exactly is it that you did there?

@Edwin Dalarzo: when I try to use Autoboxing - which i had already attempted befor posting here- (and i use Java 5 as well)- i get this error message:

C:\XXX\XXX.java:37: int cannot be dereferenced
natZahl.toString();

what does this mean?

thanx a bunch!

[Keith Lynn]

You can't call a method on a primitive.

If you want the String equivalent of an int, you can use the static method String.valueOf

[Wolfgang Obi]

thanx Keith,...

one more question though:

how do i express the following as an "If" statement in Java code ?:

if the digits entered aren't integers, then ....System.out.println("only integers pls!");

[Edwin Dalorzo]

Thanx very simple and uncomplicated...., what exactly is it that you did there?

The + operator is overloaded for String, hence you are concatenating String objects, the + operator converts the other operand to an String too.

You have to be very careful when you use that technique. Look at this examples:

 
int age = 28;
String jedi = " Obi-wan is ";
String old = " years old ";
 
String result1 = jedi + age + old; //Obi-wan is 28 years old
 
String result2 = age + old + jedi; //28 years old Obi-wan is
 
 
//Look out here
String result3 = age + age + old + jedi; //56 years old Obi-wan is
 
//Look out here too
String result4 = jedi + age + age + old; //Obi-wan is 2828 years old
 

Depending on the order of the operators you may receive different results.

As for the digits validation, well, I guess you mean checking that a String just contain valid digits.

I will assume you do not mean the same String we just converted from an int, because it would not make sense.

If you have a String and you want to check that it just have valid digits you could use the Character.isDigit() method or Integer.parse() method and check for the NumberFormatException.
[ May 06, 2006: Message edited by: Edwin Dalorzo ]

[Garrett Rowe]

If you have a String and you want to check that it just have valid digits you could use the Character.isDigit() method or Integer.parse() method and check for the NumberFormatException.

Or you could check it against a regular expression:

boolean hasAllDigits = myString.matches("\\d+");

[Wolfgang Obi]

If you have a String and you want to check that it just have valid digits you could use the Character.isDigit() method or Integer.parse() method and check for the NumberFormatException.

is this method similar to : parseInt(); ?
if not: in what class do i find "parse();" ?

thank you all so much,...very grateful!

[Tony Morris]

You can use Integer.toString instead of String.valueOf.
In fact, most implementations (at least the IBM one) calls Integer.toString from String.valueOf(int).