java Application 23 * 2
public static void main(String[] args){
System.out.println(args.length);
}
output: 7
why?
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Filburt
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Ajit Singh
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2You should take time to read this: http://stackoverflow.com/help/how-to-ask – Pradeep Simha Dec 16 '15 at 16:44
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why dont you print the args `for(String s: args){ System.out.println("arg: " + s); }` – Eric G Dec 16 '15 at 16:52
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Is that your complete command line? `23 * 2` will be parsed by most shells as arguments `"23"` `"*"` and `"2"` but your shell seems different... – billc.cn Dec 16 '15 at 16:53
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1I'm out of votes, but here's a duplicate: http://stackoverflow.com/questions/2718873/problem-of-in-command-line-argument – Sotirios Delimanolis Dec 16 '15 at 17:02
1 Answers
1
On Windows, you would get what you're expecting.
On Linux, the command shell is applying wildcard expansion, replacing the * with names of all the files in the current directory.
To prevent wildcard expansion, quote the parameter (from Stop shell wildcard character expansion?):
java Application 23 '*' 2
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1or possibly the OP wants java Application '23 * 2' - that is, a single argument to main – FredK Dec 16 '15 at 17:05