Tks Mr.miLosh & Mr.Dean,
This problem is getting clear now. However, let me please
ask one more question that still remains in me.
____________________________________________________________
Mr.miLosh advised;
>first of all, in the '-z' example u r giving a string,
>which means 'test' has something to test for. depending on
>the argument, it returns its exit code.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
No, in the '-z' example, I am NOT giveing a string.
$test -z<CR> -- NOT $test -z<space><CR>
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Mr.Dean advised;
So `test' is saying to itself "I have no arg2 to test against.
Formally,
arg2 does not exist ==> arg2 is not a string.
Hence you get an automatic "false".
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Yes, I understand why $test -n<CR> returns 0. As you mentioned,
no argment is different from no string.
But this assumptions also apply '-z' option, isn't it? If so,
$test -z<CR> should return 1 not 0, because `test' should say
to itself (as in -z case) that I have no arg to test against,
then it should return authentic `false'.
I read comments from Mr.miLosh & Mr.Dean many time and tried
test command in various patters many times. (I should have
studied English harder!!) And finally I came up with a
conclusion as Mr.which quotes;
'it is common in programms return a >0 value if a needed
argument is missing.' So test returns 0 if no arg is given
after -* option. Thus both $test -z<CR> and $test -n<CR> returns 0.
Now I forced myself to be clear on this!! But why $test<CR>
(no option, no args) returns 1 now, not 0!!
I agagin have to force myself into thinking that it's a rule.
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