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Character Literals

 
Eric Lidell
Greenhorn
Posts: 20
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Pls look at the following code
class D {
public static void main (String[] args) {
char a=\u0030;
char b=\u0031;
char c=\u0032;
char y=\u0039;
/* the values from \u0030 to \u0039 are working fine when assigned char variable without the necessary '' ( i mean '\u0031').However ,usage of \u0001 to \u0029 and \u0040 and greater give a compile time error.
what is the reason? shouldnt the assignment be illegal??? */
System.out.println(""+a+b+c+y);
}
}
Pls clarify !
 
Ernest Friedman-Hill
author and iconoclast
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Substitution of unnicode characters happens before the compiler sees the code, so when your code is compiled, it looks like this:

Because 0x30... 0x39 are the ASCII for the ten decimal digits. But if you used (for example) \u0020, that's the ASCII space character, and you can see how you'd end up with a statement like

which obviously won't compile!
Anyway, the effect of "char a = \u0030;" and "char a = '0';" are quite different; in the first case, you get a char whose numeric value is zero, while in the second case, you get one whose numeric value is 48 (30 hex) but whose character value is the digit 0. Understand? In the general case, you want to use the quotes!
 
Thomas Paul
mister krabs
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In fact, you could write your entire program in unicode characters if you wished. Would make for interesting and fun debugging sessions.
 
Eric Lidell
Greenhorn
Posts: 20
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Thanks a lot guys for the replies.They solved my problem.
Thanks ,Ernest.
 
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