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qow 10: What does var, const, ... declarations in proc/function-Parameters?

Posted on 2002-03-11
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Last Modified: 2010-04-05
hi experts,

i am starting a new quest: qow = question of the week :-)
each week i will introduce a new simple? question.

now qow 10

the first working solution will get the points (a graded).

sorry, top 15 experts, you are not allowed to solve this
q, only other can solve this question :-(

well the question is:
What does var, const, ... declarations in proc/function-Parameters, what is the effect?

an explaination is needed

let see

meikl ;-)
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Question by:kretzschmar
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6 Comments
 
LVL 2

Expert Comment

by:craig_capel
ID: 6855875
You are either bored, or have nothing to do with your points :P

0
 
LVL 27

Author Comment

by:kretzschmar
ID: 6855890
>or have nothing to do with your points :-)

yep, but thats not an answer for my q :-P
0
 
LVL 6

Expert Comment

by:zebada
ID: 6855932
Sheesh I'm bored too :)

var = pass by reference
procedure/function gets a reference to the caller's value rather than a copy. This means the procedure/function can modify the caller's value.

const = constant
procedure/function cannot modify the value that is passed to it. This also means that you cannot pass the const parameter as a var parameter to another procedure/function.


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LVL 27

Author Comment

by:kretzschmar
ID: 6855943
well, looks good, zebada

don't be bored, its just a quest.

any other declarations possible?
what the bahavaiourm if no one (var, const,..) is given?

meikl ;-)
0
 
LVL 6

Accepted Solution

by:
zebada earned 100 total points
ID: 6855972
parameter with no modifiers is passed by value.
The called function/procedure is free to modify the copy that is passed to it but it will have no effect on the value as seen by the caller.

Another option is "out".
Similar to "var" but is only used to return a value not to pass in a value.

0
 
LVL 27

Author Comment

by:kretzschmar
ID: 6855980
well, zebada,
just complete and correct :-)

thanks for participating on this quest.

next week a new ?simple question

meikl ;-)
0

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