A Question on Arrays.

  • Thread starter Thread starter Learner
  • Start date Start date
L

Learner

Hello ,
I am kinda new to VB 6.0. Is the line of code
Private strCol(62, 2) As String declares a three dimensional array?

Because in the code it is being used
as
strCol(1,0) = "x"
strCol(2,0 )= "x"
strcol(3,0) = "x"
....
strCol(1,1) = "x"
strCol(2,1) = "x"
....
strCol(1,2) = "x"
strCol(2,2) = "x"
strCol(3,2) = "x"


Please take a moment to explain.


thanks
-L
 
I am kinda new to VB 6.0.

You must be new to this newsgroup to - it's a .NET (vb2002, vb2003, and
vb2005) newgroup not a VB classic (vb1 - 6) newsgroup. You should post
VB classic questions to the microsoft.public.vb ng.

But since the question does apply to vb.Net I'll go ahead and answer.
The code you provided declares a two dimensional array. 2 dimensional
arrays are sort of like a standard database table. In this case it
would be a 62 by 2 table (which means it has 124 cells). So when you do
strCol(1,2) = "x" you are just putting the string "x" into cell (1, 2).

Does that make more sense?

Thanks,

Seth Rowe
 
Hello rowe,
Thanks for answering my question. I understand your explanation. But
the thing I don't understand is

1st set with '0'
strCol(1,0)
strCol(2,0)
....
strCol(62,0)



2nd set with '1'
strCol(1,1)
strCol(2,1)
.....
strCol(62,1)


3rd set with '2'
strCor(1,2)
strCol(2,2)
....
strCol(62,2)


is not it 62X3 = 186 as it starts with '0' ?

Please take one more minute to clarify.

thanks
-L
 
Hello rowe,
Thanks for answering my question. I understand your explanation. But
the thing I don't understand is

1st set with '0'
strCol(1,0)
strCol(2,0)
....
strCol(62,0)



2nd set with '1'
strCol(1,1)
strCol(2,1)
.....
strCol(62,1)


3rd set with '2'
strCor(1,2)
strCol(2,2)
....
strCol(62,2)


is not it 62X3 = 186 as it starts with '0' ?

Please take one more minute to clarify.

thanks
-L
 
Oops, I made a mistake there - it should be an array sized 63 x 3 (189
cells) - Arrays where revamped in .NET so that dim str(4) as string
would declare an array with four members (str(0), str(1), str(2) and
str(3)) That same statement in vb classic declares an array with 5
members (str(0), str(1), str(2), str(3) and str(4)). This is one of the
many things that changed in .Net, and is a terrific example of why the
two newgroups are seperate from each other. If this same question was
posted in the vb classic ng they would have used the vb6 definition of
an array and correctly answered your question.

Thanks,

Seth Rowe
 
This is a rather unfortunate aspect of arrays: whether they are Zero or One
based. For example, if I declare a simple array as follows:

Dim myIntegers (1),

you might think I'm declaring one integer, but no, I'm declaring 2!
myIntegers (0) and myIntegers (1). The dimension in brackets represents the
upper inclusive bound on the set 0..1.

Now, your declaration, (62, 2), means you have 3 rows ( 0..2 ) with 63
columns ( 0..62 ), so in total you have 63 * 3 "cells" (189).
 
rowe_newsgroups said:
Oops, I made a mistake there - it should be an array sized 63 x 3 (189
cells) - Arrays where revamped in .NET so that dim str(4) as string
would declare an array with four members (str(0), str(1), str(2) and
Click to expand...

That is incorrect, the VB.Net behavior is the same as VB6. Dim str(4)
declares an array with 5 members. 0, 1, 2, 3, 4.

In the beginning, VB.Net worked the way C# did, with the number
indicating the number of members, but they changed it back to the VB6
way because there were many complaints that it broke compatibility with
VB6 code.
 
In the beginning, VB.Net worked the way C# did, with the number
indicating the number of members, but they changed it back to the VB6
way because there were many complaints that it broke compatibility with
VB6 code.
Click to expand...

Thanks for the correction! Which version switched this back?

Thanks,

Seth Rowe
 
That'll teach me to believe an article about "whats new in vb.net" that
was written during beta testing :-)

Thanks,

Seth Rowe
 
Back
Top