How to append a 2d array with a 1d array in VB?
I'm having difficulties to find a way to add 1d array to a 2d array in VB. For example:
Dim arr(,) As Integer
arr = { {0, 1}, {2, 3}, {4, 5} }
arr{3} = {6, 7}
'Now arr should be = { {0, 1}, {2, 3}, {4, 5}, {6, 7} }
Note that the code above is just a demonstration on what I want to achieve, it doesn't work.
I have been trying things suggested on forum such as:
- ReDim Preserve
- Array.Resize
- Making 2d loop to copy everysingle element to a new variable and add new element then ReDim it back to arr
All I tried, but they all doesn't seem to work. At the end of the day, I'm looking for a subroutine that could append an 2d array of unknown length for example:
Dim arr(,) As Integer
append(arr, {0, 1})
append(arr, {2, 3})
'Now arr should be = {{0, 1}, {2, 3}}
arrays vb.net multidimensional-array append
add a comment |
I'm having difficulties to find a way to add 1d array to a 2d array in VB. For example:
Dim arr(,) As Integer
arr = { {0, 1}, {2, 3}, {4, 5} }
arr{3} = {6, 7}
'Now arr should be = { {0, 1}, {2, 3}, {4, 5}, {6, 7} }
Note that the code above is just a demonstration on what I want to achieve, it doesn't work.
I have been trying things suggested on forum such as:
- ReDim Preserve
- Array.Resize
- Making 2d loop to copy everysingle element to a new variable and add new element then ReDim it back to arr
All I tried, but they all doesn't seem to work. At the end of the day, I'm looking for a subroutine that could append an 2d array of unknown length for example:
Dim arr(,) As Integer
append(arr, {0, 1})
append(arr, {2, 3})
'Now arr should be = {{0, 1}, {2, 3}}
arrays vb.net multidimensional-array append
Some of you might be wondering, why I need this in a subroutine, at the end of the day. Because in my program, I will be needing to append a 2d array from different functions and modules.
– John.Apple
Nov 20 '18 at 11:40
It would appear you want aList(Of Integer())
.
– GSerg
Nov 20 '18 at 11:44
I have to this for a school project. The criteria of the project doesn't seem to allow me to use List as data structure.
– John.Apple
Nov 20 '18 at 11:59
add a comment |
I'm having difficulties to find a way to add 1d array to a 2d array in VB. For example:
Dim arr(,) As Integer
arr = { {0, 1}, {2, 3}, {4, 5} }
arr{3} = {6, 7}
'Now arr should be = { {0, 1}, {2, 3}, {4, 5}, {6, 7} }
Note that the code above is just a demonstration on what I want to achieve, it doesn't work.
I have been trying things suggested on forum such as:
- ReDim Preserve
- Array.Resize
- Making 2d loop to copy everysingle element to a new variable and add new element then ReDim it back to arr
All I tried, but they all doesn't seem to work. At the end of the day, I'm looking for a subroutine that could append an 2d array of unknown length for example:
Dim arr(,) As Integer
append(arr, {0, 1})
append(arr, {2, 3})
'Now arr should be = {{0, 1}, {2, 3}}
arrays vb.net multidimensional-array append
I'm having difficulties to find a way to add 1d array to a 2d array in VB. For example:
Dim arr(,) As Integer
arr = { {0, 1}, {2, 3}, {4, 5} }
arr{3} = {6, 7}
'Now arr should be = { {0, 1}, {2, 3}, {4, 5}, {6, 7} }
Note that the code above is just a demonstration on what I want to achieve, it doesn't work.
I have been trying things suggested on forum such as:
- ReDim Preserve
- Array.Resize
- Making 2d loop to copy everysingle element to a new variable and add new element then ReDim it back to arr
All I tried, but they all doesn't seem to work. At the end of the day, I'm looking for a subroutine that could append an 2d array of unknown length for example:
Dim arr(,) As Integer
append(arr, {0, 1})
append(arr, {2, 3})
'Now arr should be = {{0, 1}, {2, 3}}
arrays vb.net multidimensional-array append
arrays vb.net multidimensional-array append
asked Nov 20 '18 at 11:37
John.Apple
82
82
Some of you might be wondering, why I need this in a subroutine, at the end of the day. Because in my program, I will be needing to append a 2d array from different functions and modules.
– John.Apple
Nov 20 '18 at 11:40
It would appear you want aList(Of Integer())
.
– GSerg
Nov 20 '18 at 11:44
I have to this for a school project. The criteria of the project doesn't seem to allow me to use List as data structure.
– John.Apple
Nov 20 '18 at 11:59
add a comment |
Some of you might be wondering, why I need this in a subroutine, at the end of the day. Because in my program, I will be needing to append a 2d array from different functions and modules.
– John.Apple
Nov 20 '18 at 11:40
It would appear you want aList(Of Integer())
.
– GSerg
Nov 20 '18 at 11:44
I have to this for a school project. The criteria of the project doesn't seem to allow me to use List as data structure.
– John.Apple
Nov 20 '18 at 11:59
Some of you might be wondering, why I need this in a subroutine, at the end of the day. Because in my program, I will be needing to append a 2d array from different functions and modules.
– John.Apple
Nov 20 '18 at 11:40
Some of you might be wondering, why I need this in a subroutine, at the end of the day. Because in my program, I will be needing to append a 2d array from different functions and modules.
– John.Apple
Nov 20 '18 at 11:40
It would appear you want a
List(Of Integer())
.– GSerg
Nov 20 '18 at 11:44
It would appear you want a
List(Of Integer())
.– GSerg
Nov 20 '18 at 11:44
I have to this for a school project. The criteria of the project doesn't seem to allow me to use List as data structure.
– John.Apple
Nov 20 '18 at 11:59
I have to this for a school project. The criteria of the project doesn't seem to allow me to use List as data structure.
– John.Apple
Nov 20 '18 at 11:59
add a comment |
1 Answer
1
active
oldest
votes
You need to Redim Preserve
the array, but you cannot do that because it adds another column when you want another row. So you will have to define a new array and copy the data:
Public Sub Append2DRow(ByRef arr(,) As Integer, new_row(,) As Integer)
Dim new_arr(,) As Integer
ReDim new_arr(0 To UBound(arr, 1) + 1, 0 To 1)
Array.Copy(arr, new_arr, arr.LongLength)
new_arr(UBound(new_arr, 1), 0) = new_row(0, 0)
new_arr(UBound(new_arr, 1), 1) = new_row(0, 1)
arr = new_arr
End Sub
Public Sub Append1DRow(ByRef arr(,) As Integer, new_row() As Integer)
Dim new_arr(,) As Integer
ReDim new_arr(0 To UBound(arr, 1) + 1, 0 To 1)
Array.Copy(arr, new_arr, arr.LongLength)
new_arr(UBound(new_arr, 1), 0) = new_row(0)
new_arr(UBound(new_arr, 1), 1) = new_row(1)
arr = new_arr
End Sub
Dim arr(,) As Integer
arr = {{0, 1}, {2, 3}, {4, 5}}
Append2DRow(arr, {{6, 7}})
Append1DRow(arr, {6, 7})
You can avoid this problem by forcing yourself into having an array with two rows and variable number of columns, but then you won't be able to initialize it that easily and it will probably be more difficult to work with in general:
Public Sub Append2DRow(ByRef arr(,) As Integer, new_row(,) As Integer)
ReDim Preserve arr(0 To 1, 0 To UBound(arr, 2) + 1)
arr(0, UBound(arr, 2)) = new_row(0, 0)
arr(1, UBound(arr, 2)) = new_row(0, 1)
End Sub
Public Sub Append1DRow(ByRef arr(,) As Integer, new_row() As Integer)
ReDim Preserve arr(0 To 1, 0 To UBound(arr, 2) + 1)
arr(0, UBound(arr, 2)) = new_row(0)
arr(1, UBound(arr, 2)) = new_row(1)
End Sub
Dim arr(0 To 1, 0 To 2) As Integer
arr(0, 0) = 0 : arr(1, 0) = 1
arr(0, 1) = 2 : arr(1, 1) = 3
arr(0, 2) = 4 : arr(1, 2) = 5
Append2DRow(arr, {{6, 7}})
Append1DRow(arr, {6, 7})
However you should not use arrays for this to begin with. You should use lists:
Dim list = New List(Of Integer()) From {
New Integer() {0, 1},
New Integer() {2, 3},
New Integer() {4, 5}
}
list.Add({6, 7})
Thank you very much for this GSerg. Although, I'm wondering on how do i do append the arr if the arr is empty (initial appending). For example, Dim arr(,). Then immediately followed by Append1DRow(arr, {6,7})
– John.Apple
Nov 20 '18 at 12:52
@John.Apple CheckIf arr Is Nothing
in theAppend*
functions. If it is, redimnew_arr
to(0,1)
and do notArray.Copy
. If it's not, redim to the +1 dimension and copy as shown.
– GSerg
Nov 20 '18 at 13:00
add a comment |
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1 Answer
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1 Answer
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oldest
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oldest
votes
active
oldest
votes
You need to Redim Preserve
the array, but you cannot do that because it adds another column when you want another row. So you will have to define a new array and copy the data:
Public Sub Append2DRow(ByRef arr(,) As Integer, new_row(,) As Integer)
Dim new_arr(,) As Integer
ReDim new_arr(0 To UBound(arr, 1) + 1, 0 To 1)
Array.Copy(arr, new_arr, arr.LongLength)
new_arr(UBound(new_arr, 1), 0) = new_row(0, 0)
new_arr(UBound(new_arr, 1), 1) = new_row(0, 1)
arr = new_arr
End Sub
Public Sub Append1DRow(ByRef arr(,) As Integer, new_row() As Integer)
Dim new_arr(,) As Integer
ReDim new_arr(0 To UBound(arr, 1) + 1, 0 To 1)
Array.Copy(arr, new_arr, arr.LongLength)
new_arr(UBound(new_arr, 1), 0) = new_row(0)
new_arr(UBound(new_arr, 1), 1) = new_row(1)
arr = new_arr
End Sub
Dim arr(,) As Integer
arr = {{0, 1}, {2, 3}, {4, 5}}
Append2DRow(arr, {{6, 7}})
Append1DRow(arr, {6, 7})
You can avoid this problem by forcing yourself into having an array with two rows and variable number of columns, but then you won't be able to initialize it that easily and it will probably be more difficult to work with in general:
Public Sub Append2DRow(ByRef arr(,) As Integer, new_row(,) As Integer)
ReDim Preserve arr(0 To 1, 0 To UBound(arr, 2) + 1)
arr(0, UBound(arr, 2)) = new_row(0, 0)
arr(1, UBound(arr, 2)) = new_row(0, 1)
End Sub
Public Sub Append1DRow(ByRef arr(,) As Integer, new_row() As Integer)
ReDim Preserve arr(0 To 1, 0 To UBound(arr, 2) + 1)
arr(0, UBound(arr, 2)) = new_row(0)
arr(1, UBound(arr, 2)) = new_row(1)
End Sub
Dim arr(0 To 1, 0 To 2) As Integer
arr(0, 0) = 0 : arr(1, 0) = 1
arr(0, 1) = 2 : arr(1, 1) = 3
arr(0, 2) = 4 : arr(1, 2) = 5
Append2DRow(arr, {{6, 7}})
Append1DRow(arr, {6, 7})
However you should not use arrays for this to begin with. You should use lists:
Dim list = New List(Of Integer()) From {
New Integer() {0, 1},
New Integer() {2, 3},
New Integer() {4, 5}
}
list.Add({6, 7})
Thank you very much for this GSerg. Although, I'm wondering on how do i do append the arr if the arr is empty (initial appending). For example, Dim arr(,). Then immediately followed by Append1DRow(arr, {6,7})
– John.Apple
Nov 20 '18 at 12:52
@John.Apple CheckIf arr Is Nothing
in theAppend*
functions. If it is, redimnew_arr
to(0,1)
and do notArray.Copy
. If it's not, redim to the +1 dimension and copy as shown.
– GSerg
Nov 20 '18 at 13:00
add a comment |
You need to Redim Preserve
the array, but you cannot do that because it adds another column when you want another row. So you will have to define a new array and copy the data:
Public Sub Append2DRow(ByRef arr(,) As Integer, new_row(,) As Integer)
Dim new_arr(,) As Integer
ReDim new_arr(0 To UBound(arr, 1) + 1, 0 To 1)
Array.Copy(arr, new_arr, arr.LongLength)
new_arr(UBound(new_arr, 1), 0) = new_row(0, 0)
new_arr(UBound(new_arr, 1), 1) = new_row(0, 1)
arr = new_arr
End Sub
Public Sub Append1DRow(ByRef arr(,) As Integer, new_row() As Integer)
Dim new_arr(,) As Integer
ReDim new_arr(0 To UBound(arr, 1) + 1, 0 To 1)
Array.Copy(arr, new_arr, arr.LongLength)
new_arr(UBound(new_arr, 1), 0) = new_row(0)
new_arr(UBound(new_arr, 1), 1) = new_row(1)
arr = new_arr
End Sub
Dim arr(,) As Integer
arr = {{0, 1}, {2, 3}, {4, 5}}
Append2DRow(arr, {{6, 7}})
Append1DRow(arr, {6, 7})
You can avoid this problem by forcing yourself into having an array with two rows and variable number of columns, but then you won't be able to initialize it that easily and it will probably be more difficult to work with in general:
Public Sub Append2DRow(ByRef arr(,) As Integer, new_row(,) As Integer)
ReDim Preserve arr(0 To 1, 0 To UBound(arr, 2) + 1)
arr(0, UBound(arr, 2)) = new_row(0, 0)
arr(1, UBound(arr, 2)) = new_row(0, 1)
End Sub
Public Sub Append1DRow(ByRef arr(,) As Integer, new_row() As Integer)
ReDim Preserve arr(0 To 1, 0 To UBound(arr, 2) + 1)
arr(0, UBound(arr, 2)) = new_row(0)
arr(1, UBound(arr, 2)) = new_row(1)
End Sub
Dim arr(0 To 1, 0 To 2) As Integer
arr(0, 0) = 0 : arr(1, 0) = 1
arr(0, 1) = 2 : arr(1, 1) = 3
arr(0, 2) = 4 : arr(1, 2) = 5
Append2DRow(arr, {{6, 7}})
Append1DRow(arr, {6, 7})
However you should not use arrays for this to begin with. You should use lists:
Dim list = New List(Of Integer()) From {
New Integer() {0, 1},
New Integer() {2, 3},
New Integer() {4, 5}
}
list.Add({6, 7})
Thank you very much for this GSerg. Although, I'm wondering on how do i do append the arr if the arr is empty (initial appending). For example, Dim arr(,). Then immediately followed by Append1DRow(arr, {6,7})
– John.Apple
Nov 20 '18 at 12:52
@John.Apple CheckIf arr Is Nothing
in theAppend*
functions. If it is, redimnew_arr
to(0,1)
and do notArray.Copy
. If it's not, redim to the +1 dimension and copy as shown.
– GSerg
Nov 20 '18 at 13:00
add a comment |
You need to Redim Preserve
the array, but you cannot do that because it adds another column when you want another row. So you will have to define a new array and copy the data:
Public Sub Append2DRow(ByRef arr(,) As Integer, new_row(,) As Integer)
Dim new_arr(,) As Integer
ReDim new_arr(0 To UBound(arr, 1) + 1, 0 To 1)
Array.Copy(arr, new_arr, arr.LongLength)
new_arr(UBound(new_arr, 1), 0) = new_row(0, 0)
new_arr(UBound(new_arr, 1), 1) = new_row(0, 1)
arr = new_arr
End Sub
Public Sub Append1DRow(ByRef arr(,) As Integer, new_row() As Integer)
Dim new_arr(,) As Integer
ReDim new_arr(0 To UBound(arr, 1) + 1, 0 To 1)
Array.Copy(arr, new_arr, arr.LongLength)
new_arr(UBound(new_arr, 1), 0) = new_row(0)
new_arr(UBound(new_arr, 1), 1) = new_row(1)
arr = new_arr
End Sub
Dim arr(,) As Integer
arr = {{0, 1}, {2, 3}, {4, 5}}
Append2DRow(arr, {{6, 7}})
Append1DRow(arr, {6, 7})
You can avoid this problem by forcing yourself into having an array with two rows and variable number of columns, but then you won't be able to initialize it that easily and it will probably be more difficult to work with in general:
Public Sub Append2DRow(ByRef arr(,) As Integer, new_row(,) As Integer)
ReDim Preserve arr(0 To 1, 0 To UBound(arr, 2) + 1)
arr(0, UBound(arr, 2)) = new_row(0, 0)
arr(1, UBound(arr, 2)) = new_row(0, 1)
End Sub
Public Sub Append1DRow(ByRef arr(,) As Integer, new_row() As Integer)
ReDim Preserve arr(0 To 1, 0 To UBound(arr, 2) + 1)
arr(0, UBound(arr, 2)) = new_row(0)
arr(1, UBound(arr, 2)) = new_row(1)
End Sub
Dim arr(0 To 1, 0 To 2) As Integer
arr(0, 0) = 0 : arr(1, 0) = 1
arr(0, 1) = 2 : arr(1, 1) = 3
arr(0, 2) = 4 : arr(1, 2) = 5
Append2DRow(arr, {{6, 7}})
Append1DRow(arr, {6, 7})
However you should not use arrays for this to begin with. You should use lists:
Dim list = New List(Of Integer()) From {
New Integer() {0, 1},
New Integer() {2, 3},
New Integer() {4, 5}
}
list.Add({6, 7})
You need to Redim Preserve
the array, but you cannot do that because it adds another column when you want another row. So you will have to define a new array and copy the data:
Public Sub Append2DRow(ByRef arr(,) As Integer, new_row(,) As Integer)
Dim new_arr(,) As Integer
ReDim new_arr(0 To UBound(arr, 1) + 1, 0 To 1)
Array.Copy(arr, new_arr, arr.LongLength)
new_arr(UBound(new_arr, 1), 0) = new_row(0, 0)
new_arr(UBound(new_arr, 1), 1) = new_row(0, 1)
arr = new_arr
End Sub
Public Sub Append1DRow(ByRef arr(,) As Integer, new_row() As Integer)
Dim new_arr(,) As Integer
ReDim new_arr(0 To UBound(arr, 1) + 1, 0 To 1)
Array.Copy(arr, new_arr, arr.LongLength)
new_arr(UBound(new_arr, 1), 0) = new_row(0)
new_arr(UBound(new_arr, 1), 1) = new_row(1)
arr = new_arr
End Sub
Dim arr(,) As Integer
arr = {{0, 1}, {2, 3}, {4, 5}}
Append2DRow(arr, {{6, 7}})
Append1DRow(arr, {6, 7})
You can avoid this problem by forcing yourself into having an array with two rows and variable number of columns, but then you won't be able to initialize it that easily and it will probably be more difficult to work with in general:
Public Sub Append2DRow(ByRef arr(,) As Integer, new_row(,) As Integer)
ReDim Preserve arr(0 To 1, 0 To UBound(arr, 2) + 1)
arr(0, UBound(arr, 2)) = new_row(0, 0)
arr(1, UBound(arr, 2)) = new_row(0, 1)
End Sub
Public Sub Append1DRow(ByRef arr(,) As Integer, new_row() As Integer)
ReDim Preserve arr(0 To 1, 0 To UBound(arr, 2) + 1)
arr(0, UBound(arr, 2)) = new_row(0)
arr(1, UBound(arr, 2)) = new_row(1)
End Sub
Dim arr(0 To 1, 0 To 2) As Integer
arr(0, 0) = 0 : arr(1, 0) = 1
arr(0, 1) = 2 : arr(1, 1) = 3
arr(0, 2) = 4 : arr(1, 2) = 5
Append2DRow(arr, {{6, 7}})
Append1DRow(arr, {6, 7})
However you should not use arrays for this to begin with. You should use lists:
Dim list = New List(Of Integer()) From {
New Integer() {0, 1},
New Integer() {2, 3},
New Integer() {4, 5}
}
list.Add({6, 7})
answered Nov 20 '18 at 12:26
GSerg
58.9k14101219
58.9k14101219
Thank you very much for this GSerg. Although, I'm wondering on how do i do append the arr if the arr is empty (initial appending). For example, Dim arr(,). Then immediately followed by Append1DRow(arr, {6,7})
– John.Apple
Nov 20 '18 at 12:52
@John.Apple CheckIf arr Is Nothing
in theAppend*
functions. If it is, redimnew_arr
to(0,1)
and do notArray.Copy
. If it's not, redim to the +1 dimension and copy as shown.
– GSerg
Nov 20 '18 at 13:00
add a comment |
Thank you very much for this GSerg. Although, I'm wondering on how do i do append the arr if the arr is empty (initial appending). For example, Dim arr(,). Then immediately followed by Append1DRow(arr, {6,7})
– John.Apple
Nov 20 '18 at 12:52
@John.Apple CheckIf arr Is Nothing
in theAppend*
functions. If it is, redimnew_arr
to(0,1)
and do notArray.Copy
. If it's not, redim to the +1 dimension and copy as shown.
– GSerg
Nov 20 '18 at 13:00
Thank you very much for this GSerg. Although, I'm wondering on how do i do append the arr if the arr is empty (initial appending). For example, Dim arr(,). Then immediately followed by Append1DRow(arr, {6,7})
– John.Apple
Nov 20 '18 at 12:52
Thank you very much for this GSerg. Although, I'm wondering on how do i do append the arr if the arr is empty (initial appending). For example, Dim arr(,). Then immediately followed by Append1DRow(arr, {6,7})
– John.Apple
Nov 20 '18 at 12:52
@John.Apple Check
If arr Is Nothing
in the Append*
functions. If it is, redim new_arr
to (0,1)
and do not Array.Copy
. If it's not, redim to the +1 dimension and copy as shown.– GSerg
Nov 20 '18 at 13:00
@John.Apple Check
If arr Is Nothing
in the Append*
functions. If it is, redim new_arr
to (0,1)
and do not Array.Copy
. If it's not, redim to the +1 dimension and copy as shown.– GSerg
Nov 20 '18 at 13:00
add a comment |
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Some of you might be wondering, why I need this in a subroutine, at the end of the day. Because in my program, I will be needing to append a 2d array from different functions and modules.
– John.Apple
Nov 20 '18 at 11:40
It would appear you want a
List(Of Integer())
.– GSerg
Nov 20 '18 at 11:44
I have to this for a school project. The criteria of the project doesn't seem to allow me to use List as data structure.
– John.Apple
Nov 20 '18 at 11:59