#35 scalar definition

closed-fixed
5
2008-01-31
2008-01-29
futurdorko
No

The manual (p. 58) claims it's permitted to omit any of the 3 extra parameters of a scalar, but taking out the default one something goes wrong.

The following example does NOT produces an error, while it supposed to.
<i>
function foo( scalar a[0:2:])

print a

end function

nulldata 100

a = 3000

foo(a)
</i>

Cheers

Discussion

  • Allin Cottrell
    Allin Cottrell
    2008-01-31

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    user_id=330339
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    You're right. Fixed in CVS.

     
  • Allin Cottrell
    Allin Cottrell
    2008-01-31

    • labels: --> Script commands
    • assigned_to: nobody --> allin
    • status: open --> open-fixed
     
  • Allin Cottrell
    Allin Cottrell
    2008-01-31

    • status: open-fixed --> closed-fixed