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Hi!!

How can I convert an String into an Integer? I'm trying to use
'parseInt' but I only get a "java.lang.NumberFormatException".

String s = '\n';
Integer i = Integer.parseInt(s); // i should be 10

I also tried with Integer.valueof(s) with no success. It works for a
char, but not for a String. How can I solve it??

---
Exception in thread "main" java.lang.NumberFormatException: For input
string: "hello"
at java.lang.NumberFormatException.forInputString(Unk nown Source)
at java.lang.Integer.parseInt(Unknown Source)
at java.lang.Integer.parseInt(Unknown Source)
---

Thanks in advance!

Lew wrote:

> Integer i = '\n';
>
> is what you want.
>

Thanks for answering so quickly!

The thing is that I have a String var with characters such as "0", "a",
"\n"... I use String.charAt(0) for getting the first character of the
string. It works nice when I'm using a normal character ('0', 'a'...)
but it doesnt with escape characters (it gets '\' instead of '\n').

So the question is: how can convert a string "\n" into an integer?
Or how can I convert "\n" into a char for then converting it to an integer??

Thanks in advance.

adam wrote:
> Lew wrote:
>
>> Integer i = '\n';
>>
>> is what you want.
>>
>
> Thanks for answering so quickly!
>
> The thing is that I have a String var with characters such as "0", "a",
> "\n"... I use String.charAt(0) for getting the first character of the
> string. It works nice when I'm using a normal character ('0', 'a'...)
> but it doesnt with escape characters (it gets '\' instead of '\n').
>
> So the question is: how can convert a string "\n" into an integer?
> Or how can I convert "\n" into a char for then converting it to an
> integer??

\n is an escape sequence for string and character literals. If read
from a file or the keyboard, the IO system does not assume that you
really meant a \ to mean something else, it just reads what is there.
If you want a newline in a file or at the keyboard, hit Enter.

I don't know of Java class that will parse string literal escapes for
you. Considering there's only a few of them, doing it yourself should
not be hard.

adam wrote:
> Lew wrote:
>
>> Integer i = '\n';
>>
>> is what you want.
>>
>
> Thanks for answering so quickly!
>
> The thing is that I have a String var with characters such as "0", "a",
> "\n"... I use String.charAt(0) for getting the first character of the
> string. It works nice when I'm using a normal character ('0', 'a'...)
> but it doesnt with escape characters (it gets '\' instead of '\n').
>
> So the question is: how can convert a string "\n" into an integer?
> Or how can I convert "\n" into a char for then converting it to an
> integer??
>
> Thanks in advance.
I disagree:
<sscce>
public class Main {
public static void main(String[] args) {
String string = "Test\n";
for (int i = 0; i < string.length(); ++i) {
char c = string.charAt(i);
System.out.print(c);
System.out.println(" - " + (int)c);
}
}
}
</sscce>
<output>
T - 84
e - 101
s - 115
t - 116

- 10
</output>

--
Daniel Pitts' Tech Blog: <http://virtualinfinity.net/wordpress/>

On Sat, 21 Jun 2008 15:41:02 +0200, adam <> wrote,
quoted or indirectly quoted someone who said :

>String s = '\n';
>Integer i = Integer.parseInt(s); // i should be 10

if you simply want the 10, there is no need to parse, namely convert a
string a character digits to binary. It is already in binary.

parseInt wants something like this:

>String s = "10";
>Integer i = Integer.parseInt(s); // i should be 10


Just do

int lf = '\n';
--

Roedy Green Canadian Mind Products
The Java Glossary
http://mindprod.com