A Question on Arrays.

13 Nov 2006 3:40 PM

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

13 Nov 2006 3:51 PM

rowe_newsgroups

> 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

Learner wrote:
Show quoteHide quote

> 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

13 Nov 2006 4:09 PM

Learner

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

rowe_newsgroups wrote:
Show quoteHide quote

> > 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
>
>
> Learner wrote:
> > 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

13 Nov 2006 4:25 PM

rowe_newsgroups

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

Learner wrote:
Show quoteHide quote

> 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
>
>
> rowe_newsgroups wrote:
> > >    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
> >
> >
> > Learner wrote:
> > > 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

13 Nov 2006 6:40 PM

Chris Dunaway

rowe_newsgroups wrote:

> 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

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.

13 Nov 2006 7:09 PM

rowe_newsgroups

> 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.

Thanks for the correction! Which version switched this back?

Thanks,

Seth Rowe

Chris Dunaway wrote:
Show quoteHide quote

> rowe_newsgroups wrote:
> > 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
>
> 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.

13 Nov 2006 7:21 PM

Chris Dunaway

rowe_newsgroups wrote:

> > 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.
>
> Thanks for the correction! Which version switched this back?
>

Version 1.0. The new syntax never made it out of beta IIRC.

13 Nov 2006 7:53 PM

rowe_newsgroups

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

Thanks,

Seth Rowe

Chris Dunaway wrote:
Show quoteHide quote

> rowe_newsgroups wrote:
> > > 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.
> >
> > Thanks for the correction! Which version switched this back?
> >
>
> Version 1.0. The new syntax never made it out of beta IIRC.

13 Nov 2006 4:14 PM

Learner

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

rowe_newsgroups wrote:
Show quoteHide quote

> > 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
>
>
> Learner wrote:
> > 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

13 Nov 2006 4:35 PM

Robinson

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).

Show quoteHide quote

"Learner" <pra***@gmail.com> wrote in message
news:1163434486.240149.94660@m73g2000cwd.googlegroups.com...
> 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
>
>
> rowe_newsgroups wrote:
>> > 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
>>
>>
>> Learner wrote:
>> > 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
>