Vector Dimensions

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Re: Vector Dimensions

 
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  #11
Sep 30th, 2008
Correct me if I'm wrong, but doesn't the cavet ^ symbol in C++.NET mean that the type is actually a pointer?

c++ Syntax (Toggle Plain Text) 
  1. List1.Add(Vec1());
looks fishy... Vec1 is a pointer-type, so doesn't that mean you need to call gcnew to properly instantiate that Vec? O_O

c++ Syntax (Toggle Plain Text) 
  1. List1.Add(gcnew Vec1());
I'd assume.
Last edited by Alex Edwards; Sep 30th, 2008 at 12:07 am.
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Re: Vector Dimensions

 
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  #12
Sep 30th, 2008
When using pointers in .NET, you do use * and & to address pointers so the cavet ^ symbol should only meen a type as int and double I beleive like string for the std:: and then String^ for managed.
If I put gcnew before Vec1 in the code below the compiler says this message:

'Vec1' : cannot use this type as argument of 'gcnew'

typedef List<String^>^ Vec1;
typedef List<Vec1> Vec2;

Vec2 List1;

for( int j = 0; j < 1000; j++ )
{	

	List1.Add(gcnew Vec1());

	for( int i = 0; i < 1000; i++ )
	{
		List1[j]->Add("0");
	}

}
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Re: Vector Dimensions

 
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  #13
Sep 30th, 2008
I think you are after ...
C++ Syntax (Toggle Plain Text) 
  1. typedef List<String^> Vec1;
  2. typedef List<Vec1^> Vec2;
  3.  
  4. Vec2 List1;
  5.  
  6. for( int j = 0; j < 1000; j++ )
  7. {	
  8. 	List1.Add(gcnew Vec1());
  9.  
  10. 	for( int i = 0; i < 1000; i++ )
  11. 	{
  12. 		List1[j]->Add("0");
  13. 	}
  14. }
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Re: Vector Dimensions

 
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  #14
Sep 30th, 2008
Mitrmkar,

Thank you! Now it works...
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Re: Vector Dimensions

 
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  #15
Oct 2nd, 2008
When compiling the below code I will have a compileerror that says:
String3D : illegal use of this type as an expression
String2D : illegal use of this type as an expression
String1D : illegal use of this type as an expression

I wonder if I doing this correct when trying to iterate through all 3 Dimension with .Count ?


typedef List<String^> String1D;
typedef List<String1D^> String2D;
typedef List<String2D^> String3D;



	String3D Dimensions;	
	int count1 = 0;
	int count2 = -1;


for( int i = 0; i < 5; i++ )
{

	if( i == 0 )
	{
	          Dimensions.Add(gcnew String2D()); 
	}
	if( i >= 0 && i <= 4 )
	{
		count2 = count2 + 1;
		Dimensions[count1]->Add(gcnew String1D()); 
	}
	Dimensions[count1][count2]->Add("TextString");
} 


		
for(int h = 0; h < String3D.Count; ++h)
{
for(int j = 0; j < String2D.Count; ++j)
{
for(int k = 0; k < String1D.Count; ++k)
{
}}}
Last edited by Jennifer84; Oct 2nd, 2008 at 4:01 pm.
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Re: Vector Dimensions

 
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  #16
Oct 2nd, 2008
I found out that "()" was needed. When iterating through the dimensions to show the "TextString" that was put into the List(Dimensions), only 1 MessageBox will appear.
As I have put "TextString" to 5 elements it should be 5 Times.

I am putting the string to these elements:
Dimensions[0][0][0];
Dimensions[0][1][0];
Dimensions[0][2][0];
Dimensions[0][3][0];
Dimensions[0][4][0];


typedef List<String^> String1D;
typedef List<String1D^> String2D;
typedef List<String2D^> String3D;



	String3D Dimensions;	
	int count1 = 0;
	int count2 = -1;


for( int i = 0; i < 5; i++ )
{

	if( i == 0 )
	{
	          Dimensions.Add(gcnew String2D()); 
	}
	if( i >= 0 && i <= 4 )
	{
		count2 = count2 + 1;
		Dimensions[count1]->Add(gcnew String1D()); 
	}
	Dimensions[count1][count2]->Add("TextString");
} 


		
for(int h = 0; h < String3D().Count + 1; ++h)
{
for(int j = 0; j < String2D().Count + 1; ++j)
{
for(int k = 0; k < String1D().Count + 1; ++k)
{

MessageBox::Show(Dimensions[h][j][k]);

}


}}
Last edited by Jennifer84; Oct 2nd, 2008 at 4:31 pm.
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Re: Vector Dimensions

 
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  #17
Oct 2nd, 2008
What I try to find, is the equavilent to the vector ::iterator for the first code.
The first code iterates each Dimension from .begin() to the .end() and then continues
to next dimension for the 3 Dimensions.

C++ Syntax (Toggle Plain Text) 
  1. typedef std::vector<string> String1D;
  2. typedef std::vector<String1D> String2D;
  3. typedef std::vector<String2D> String3D;
  4.  
  5. String3D Dimensions;	
  6.  
  7. for (String3D::iterator i = Dimensions.begin(); i != Dimensions.end(); ++i) 
  8. {
  9. for (String2D::iterator j = i->begin(); j != i->end(); ++j)			
  10. {
  11. for (String1D::iterator k = j->begin(); k != j->end(); ++k)			
  12. {}}}

As all these Dimensions below for "List" have been ->Add and therefore have different many elements for each Dimension, I am looking for the same(equavilent) method to Iterate each Dimension to the end as the above approach for the std::vector.
Something might be missing in the below code.
C++ Syntax (Toggle Plain Text) 
  1. typedef List<String^> String1D;
  2. typedef List<String1D^> String2D;
  3. typedef List<String2D^> String3D;
  4.  
  5. String3D Dimensions;	
  6.  
  7. for(int i = 0; i < String3D().Count + 1; ++i)
  8. {
  9. for(int j = 0; j < String2D().Count + 1; ++j)
  10. {
  11. for(int k = 0; k < String1D().Count + 1; ++k)
  12. {}}}
Last edited by Jennifer84; Oct 2nd, 2008 at 7:03 pm.
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Re: Vector Dimensions

 
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  #18
Oct 3rd, 2008
In some way it is needed to refer to the previous vectordimension.
The problem is that I dont know how you refer to the previous dimension beginning and end and set this to some sort of an iterator as this is put ?

String2D::iterator j = i->begin(); j != i->end(); ++j

For the List, I cant find either a begin or end member or iterator.
The below code could be a step closer perheps.
Any idéas or tips would be of help. I am quite stuck.

C++ Syntax (Toggle Plain Text) 
  1. for(int i = 0; i < Dimensions.Count; ++i)
  2. {
  3. for(int j = 0; j < String3D().Count; ++j)
  4. {
  5. for(int k = 0; k < String2D().Count; ++k)
  6. {	
  7. }}}
Last edited by Jennifer84; Oct 3rd, 2008 at 9:40 am.
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Re: Vector Dimensions

 
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  #19
Oct 3rd, 2008
About this thread starting question: look at my posts in solved threads (two month ago):
http://www.daniweb.com/forums/thread136351.html
http://www.daniweb.com/forums/thread137814.html
Last edited by ArkM; Oct 3rd, 2008 at 12:44 pm.
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Re: Vector Dimensions

 
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  #20
Oct 3rd, 2008
Yes, this was the thread I was reading on but I missed it and as I saw it was as simple as this. I was imagining me something different.
It helped alot. Great !

C++ Syntax (Toggle Plain Text) 
  1. for(int i = 0; i < Dimensions.Count; ++i)
  2. {
  3. for(int j = 0; j < Dimensions[i]->Count; ++j)
  4. {
  5. for(int k = 0; k < Dimensions[i][j]->Count; ++k)
  6. {
  7.  
  8. 	MessageBox::Show(Dimensions[i][j][k]);
  9.  
  10. }
  11.  
  12.  
  13. }
  14. }

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Originally Posted by ArkM View Post
About this thread starting question: look at my posts in solved threads (two month ago):
http://www.daniweb.com/forums/thread136351.html
http://www.daniweb.com/forums/thread137814.html
Last edited by Jennifer84; Oct 3rd, 2008 at 1:44 pm.
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