Everything is fine when I declare
String a;
but it says Syntax error on token "1", invalid VariableDeclaratorId when I do this
String 1;
I guess it must be very trivial but I don't get it.
views:
204answers:
8
+3
A:
The parser can't distinguish it from the int literal, so it's disallowed.
Ignacio Vazquez-Abrams
2010-01-27 11:12:36
Why char and int literal treated so differently ?
Ravi Gupta
2010-01-27 11:13:59
Char literals have quotes around them.
Ignacio Vazquez-Abrams
2010-01-27 11:18:44
Because one of them is a name of a variable, the other one is a value. If you couldn't distinguish them by their naming conventions, how else should the compiler find out if it's dealing with a variable name or a value?
Kosi2801
2010-01-27 11:21:01
+5
A:
The rules for identifiers in the Java language specification state that you cannot start an identifier with a number.
Mark
2010-01-27 11:14:06
+23
A:
Well, first of all, it's because it's written in the Java language Specification.
But, maybe that this example will help you more:
String 1 = "toto"
System.out.println(1 + 2)
What should be the output?
gizmo
2010-01-27 11:14:20
+7
A:
Because 1 is also a value (which, among others, you can assign) the parser cannot know what you mean.
Consider the following snippet:
int 1 = 10;
int a = 1; // what is the value of a ? 1 or 10?
Therefore, starting a variable name with a number is dissallowed. You can use _1 instead if you really want (note that it is difficult to read though)
laura
2010-01-27 11:14:54
Suffixes to variables make a distinction between different value types. For examples 1L means that this is a value of the type Long, not integer. Or 0x... means that there is a hex-value following.
Kosi2801
2010-01-27 11:19:35
+2
A:
Not only the parser would have a great deal of effort distinguishing between an int literal and a variable (if not totally impossible) but you could end up with strange situations like:
int 1 = 999;
System.out.println(1);
// output: 1 or 999
Basically this doesn't make much sense.
bruno conde
2010-01-27 11:18:33
Yes, it does. here is the code: public static void main(String args[]) { String 1 = "one"; }it does compile on jdk1.1.3
walter
2010-01-27 11:25:00
That's not true. This was never allowed in Java and also the 1.1.3 compiler would reject this. Altough you could always have a variable named "1" in the class file (still in the lastest Java 1.6), but not in the source.
x4u
2010-01-27 12:55:29
A:
Because 1 is a value. What will someone make of this :
String 1 = "6";
String s = 1 + "00"; // With value "100" or "600"?
Similarly, true, false, null cannot be variable names.
fastcodejava
2010-01-27 11:24:39
A:
It works on my machine too: public static void main(String args[]) { String l = "one"; } I am on jdk1.7
javaExpert
2010-01-27 18:00:27
Yes, the lowercase letter "L" is a valid identifier in Java, as your compiler says. But the question wasn't about that. It was about the digit "1", which is not a valid identifier. Try it...
Paul Clapham
2010-01-27 22:31:37