hi all,
I have a String s1 = "0123456789" and I want to deal with it as an int. That is, I want to multiply it by some numbers. I tried to cast it but it didn't work.
please don't tell me to difine it as int num1 = 0123456789; because the varialble I want to deal with is a string that comes to my application from another application as a string and I can't do anything about it.
regards
Check what the Integer class can do for you:
http://java.sun.com/j2se/1.5.0/docs/api/java/lang/Integer.html
Typing your subject line into the "Search Forums" text field yields:
http://onesearch.sun.com/search/onesearch/index.jsp?qt=Converting+a+String+to+an+int+&subCat=siteforumid%3Ajava31&site=dev&dftab=siteforumid%3Ajava31&chooseCat=javaall&col=developer-forums
Are any of those posts helpful?
Good luck
Lee
> please don't tell me to define it as > int num1 = 0123456789;
> because the variable I want to deal with
> is a string that comes to my application from another
> application as a string and I can't do anything about it.
That would not compile anyway.
The leading zero makes it an octal literal and only digits [0-7] are allowed.
TimTheEnchantor, thanks for your response,
I tried to do the following:
String s1 = "23";
int a = 0;
Integer e = new Integer(s1);
a = e * 5;
it says cannot convert from Integer to int and it also says that the operators (*,/,<,>,......) are not defined for the type integer.
any ideas ?
You're almost there. Don't use the constructor, search for a static method that takes a string and returns the parsed int.
> You're almost there. Don't use the constructor,
> search for a static method that takes a string and
> returns the parsed int.
Just to add to JoachimSauer's suggestion: read reply #1.
~
public class StringToInt {
public static void main (String[] args) {
String s1 = "0123456789" ;
try {
int i = Integer.parseInt(s1.trim());
System.out.println("int i = " + i);
} catch (NumberFormatException nfe) {
System.out.println("NumberFormatException: " + nfe.getMessage());
}
}
}
Is there an advantage to using Integer.parseInt() over Integer.decode()? It seems as if decode() calls parseInt() anyway.
> Is there an advantage to using Integer.parseInt()
> over Integer.decode()? It seems as if decode() calls
> parseInt() anyway.
They have different specs. decode accepts hex, decimal and octal formats, while parseInt accepts only decimal.
thanks guys, I did the following and it worked:
String s = "123";
Integer i = new Integer(s);
int a = s.intValue();
thanks :)
> They have different specs. decode accepts hex,
> decimal and octal formats, while parseInt accepts
> only decimal.
Not quite what you were talking about, but parseInt() is overloaded to handle more than just decimal...
http://java.sun.com/javase/6/docs/api/java/lang/Integer.html#parseInt(java.lang.String,%20int)
~
> String s = "123";
> Integer i = new Integer(s);
> int a = s.intValue();
If you feel like doing that more simply, follow the suggestions given:
String s = "123";
int a = Integer.parseInt(s);
> String s = "123";
> Integer i = new Integer(s);
>
> You can do that?
Yes, for as long as the Integer class has existed.
You can also do...
int i = new Integer(s);
...with Java 5 and later.
~
