Question for anybody who knows Java - Democratic Underground

billyskank

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Thu Aug-31-06 04:29 AM
Original message

Question for anybody who knows Java

Okay, I know this must be the Dumbest Question Ever. But how do you know how many arguments have been passed to your Java program?

Like, in C I declare my main() function like this:

void main(int argc, char *argv)

and argc tells me how many values there are in the argv array. But in Java I have to do this:

public static void main(String[] args)

so how on earth do I know how many values there are in the args array?

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Fredda Weinberg

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Thu Aug-31-06 05:13 AM
Response to Original message

1. You've declared it

Then you'll initialize with the "new" operator or just assign values

names = new String<100>;

or

String[] names = {"Fred", "Wilma", "Pebbles"};

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billyskank

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Thu Aug-31-06 05:20 AM
Response to Reply #1

2. I don't want to assign values to it

it already has values assigned to it, which are the arguments passed to the program from the command line. So how do I iterate through them - how do I know how many there are?

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Fredda Weinberg

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Thu Aug-31-06 08:19 PM
Response to Reply #2

4. To continue our names() example

for(int i=0; i<names.length; i++){}

Obviously, you can do what you like in the curly braces. Reference each item with

names

Ain't Java terse?

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Commie Pinko Dirtbag

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Thu Aug-31-06 12:26 PM
Response to Original message

3. C man here, but I'd bet that array type has some "size" property

Like args.length() or something.

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aePrime

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Thu Aug-31-06 09:50 PM
Response to Original message

5. Somebody's mentioned the first...

Edited on Thu Aug-31-06 09:53 PM by aePrime

public class Array
{
public static void main(String[] args)
{
for (int i = 0; i < args.length; ++i) {
System.out.println(args);
}

// Java 5.0
for (String element : args) {
System.out.println(element);
}
}
}

The first will work with any Java version. The second works only in Java 5. The arrays in Java have a length, similar to C++'s STL std::vector (I actually spend all of my time writing C++, but am familiar with Java).

I noticed that the first guy who wrote the loop used "i++". I'm in the habit of writing "++i". Because Java doesn't have operator overloading, it probably doesn't matter, but in C++ "++i" is more efficient than "i++".

(Updated to make the formatting better. DU insists on italiasizing parts of this message.)

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aePrime

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Thu Aug-31-06 09:54 PM
Response to Original message

6. Also

Be aware that unlike C, Java doesn't store the executable name in the first array slot. All values in the array are arguments.

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Sat Apr 20th 2024, 09:04 AM
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