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ArrayIndexOutOfBounds exception

Fantine Ponter
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Joined: Aug 11, 2008
Posts: 30
Code Magnets in Head First Java, page 351:
The following code (Exercise solution straight from the book) compiles, but when I run it, I get the above exception on main because of the statement " String test = args[0];"

When I comment out that statement and uncomment my own statement just below that (String test = "no";) , I get no runtime error.

Please would someone explain what the statement, String test = args[0];
means, and why I get a runtime error. thanks in advance.

[edit]Disable smilies. CR[/edit]
[ December 05, 2008: Message edited by: Campbell Ritchie ]
Paul Clapham
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Joined: Oct 14, 2005
Posts: 18541
    
    8

It gets the first of the parameters that you typed on the command line after the class name.

If you didn't type any parameters, then you get the error that you observe.
marc weber
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Joined: Aug 31, 2004
Posts: 11343

The String array in main contains the arguments you supply at runtime. If you do not supply any arguments, then this array will have a length of zero, and trying to access its elements will result in an exception.

To supply arguments at runtime, enter them on the command line after the class name. For example...

java ShowMyArgs here are some args
[ December 05, 2008: Message edited by: marc weber ]

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Fantine Ponter
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Joined: Aug 11, 2008
Posts: 30
Thanks Paul and Marc.

I'll try to be more observant in future!
 
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