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#define R_GLDRAWELEMENTS 0x5ED02FD4
WINAPI _R_glDrawElements(GLenum mode, GLsizei count, GLenum type, const GLvoid *indices);
WINAPI (*R_glDrawElements)(GLenum , GLsizei , GLenum , const GLvoid *) = (void (__stdcall*) (unsigned int,int,unsigned int,void const *)) R_GLDRAWELEMENTS;
WINAPI _R_glDrawElements(GLenum mode, GLsizei count, GLenum type, const GLvoid *indices)
{
_R_glDrawElements( mode, count, type, indices);
}
The 3 line of code gives a bad error error C2059: syntax error : '('
And i dont know what is wrong with it the brackets seems to be all in place....
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I'm not 100% sure how you didn't get other compiler errors, to be honest. I get errors where you've used WINAPI, as that doesn't (on my Windows installation) specify a return type - I have #define WINAPI __stdcall - just a calling convention.
Also - in the body of _R_glDrawElements, you call _R_glDrawElements recursively - that'll cause a stack overflow. Did you really mean R_glDrawElements??
Anyway - try this (modulo changing the body of _R_glDrawElements if I was correct above)
#define R_GLDRAWELEMENTS 0x5ED02FD4
void WINAPI _R_glDrawElements(GLenum mode, GLsizei count, GLenum type, const GLvoid *indices);
void (WINAPI* R_glDrawElements)(GLenum , GLsizei , GLenum , const GLvoid *) = (void (__stdcall*) (unsigned int,int,unsigned int,void const *)) R_GLDRAWELEMENTS;
void WINAPI _R_glDrawElements(GLenum mode, GLsizei count, GLenum type, const GLvoid *indices)
{
_R_glDrawElements( mode, count, type, indices);
}
Java, Basic, who cares - it's all a bunch of tree-hugging hippy cr*p
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In addition to Stuart's reply, it is recommended to type define function pointers.
It makes the code all the more readable.
typedef void (WINAPI* PFN_GLDRAWELEMENTS)(GLenum , GLsizei , GLenum , const GLvoid *);
PFN_GLDRAWELEMENTS R_glDrawElements = (PFN_GLDRAWELEMENTS)R_GLDRAWELEMENTS;
«_Superman_»
I love work. It gives me something to do between weekends.
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I tryed your methods but it dosent work ....
typedef VOID (WINAPI* R_glDrawElements_) (GLenum , GLsizei , GLenum , const GLvoid *);
R_glDrawElements_ pR_glDrawElements = 0x5ED02FD4;
And doing this i get error
C:\Documents and Settings\ivor\Desktop\test5\Client.cpp(23) : error C2440: 'initializing' : cannot convert from 'const int' to 'void (__stdcall *)(unsigned int,int,unsigned int,const void *)'
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nah1337 wrote: R_glDrawElements_ pR_glDrawElements = 0x5ED02FD4;
You need the typecast here.
R_glDrawElements_ pR_glDrawElements = (R_glDrawElements_)0x5ED02FD4;
«_Superman_»
I love work. It gives me something to do between weekends.
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Hello again thnx for previous replay well i still struggling with it and now i dosent just understand why i get the error i dont change anything after typedef ....
typedef VOID (WINAPI* R_glDrawElements_) (GLenum , GLsizei , GLenum , const GLvoid *);
R_glDrawElements_ pR_glDrawElements = (R_glDrawElements_)0x5ED02FD4;
VOID WINAPI R_glDrawElements_(GLenum mode, GLsizei count, GLenum type, const GLvoid *indices)
{
R_glDrawElements_( mode, count, type, indices);
}
C:\Documents and Settings\ivor\Desktop\test5\Client.cpp(56) : error C2377: 'R_glDrawElements_' : redefinition; typedef cannot be overloaded with any other symbol
C:\Documents and Settings\ivor\Desktop\test5\Client.cpp(27) : see declaration of 'R_glDrawElements_'
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Hi
When I tried to delete a class instance, I got the error message"HEAP: Heap block at 003BCB48 modified at 003BCC6C past requested size of 11c".
What is wrong?
Best regards,
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You allocated a block of memory, some 284 bytes in size, and then accessed something just beyond the end of the chunk that was given to you, at offset 292.
This snippet is likely to cause a similar error:
char* p=malloc(284);
p[292]=0;
Luc Pattyn [Forum Guidelines] [My Articles]
The quality and detail of your question reflects on the effectiveness of the help you are likely to get.
Show formatted code inside PRE tags, and give clear symptoms when describing a problem.
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You wrote past the end of the object somehow. The heap manager in debug mode puts sentinel values before and after allocations and checks these when you deallocate the memory. If the sentinel values have changed, it means you've written into memory you shouldn't have done.
Java, Basic, who cares - it's all a bunch of tree-hugging hippy cr*p
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To expand on what I said earlier, the problem will typically show when the memory manager checks the borders of yout allocation, so if your code would look like:
CMyclass * pCMyclass = new CMyclass ();
...
delete pCMyclass;
then the problem would occur when executing the delete statement, although the error could be anywhere in the CMyclass object.
Luc Pattyn [Forum Guidelines] [My Articles]
The quality and detail of your question reflects on the effectiveness of the help you are likely to get.
Show formatted code inside PRE tags, and give clear symptoms when describing a problem.
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Please tell me how function call is resolve, how vptr works in this case CODE:
class base
{
public:
virtual void f1()
{
cout<<"base f1";
}
virtual void f2()
{
cout<<"base f2";
}
};
class d1: public base
{
public:
virtual void f1()
{
cout<<"d1's f1";
}
virtual void f2()
{
cout<<"d1's f2";
}
};
class d2: virtual public base,virtual public d1
{
public:
d2()
{
cout<<"d2 class construxtor\n";
}
~d2()
{
cout<<" d2 class destructor\n";
}
};
void main()
{
clrscr();
base* obj= new d2;
obj->f1();
obj->f2();
}
modified on Sunday, July 12, 2009 2:46 PM
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vptrs don't come into it - you have no virtual functions, so therefore there will be no vtable.
If the compiler was able to construct your base pointer obj (it can't because it inherits from base twice - which one should it upcast through?), there would be no runtime resolution of f1 and f2; they're non-virtual functions, so they will be resolved using the static type of obj, i.e. base. That means that the program (if it could compile) would output base f1base f2
Java, Basic, who cares - it's all a bunch of tree-hugging hippy cr*p
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Actually the code will (and does) compile. Note that base is inherited virtually (note the use of virtual public base in the declaration of class d2). This ensures that only one version of the base class will be included in d2
However, you are completely correct in that there is no vtable, and the output wil be base f1base f2.
Graham
Librarians rule, Ook!
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Better tell that to gcc, then - it gives this warning and error:
a.cpp:30: warning: direct base ‘base’ inaccessible in ‘d2’ due to ambiguity
a.cpp: In function ‘int main()’:
a.cpp:46: error: ‘base’ is an ambiguous base of ‘d2’
VC++ 2008 agrees
a.cpp(30) : error C2584: 'd2' : direct base 'base' is inaccessible; already a base of 'd1'
a.cpp(5) : see declaration of 'base'
a.cpp(16) : see declaration of 'd1'
The virtual qualifier needs to be applied to all inheritance instances of the shared base, so you need this code for it to compile (note virtual wherever 'base' is inherited):
class base
{
public:
void f1()
{
cout<<"base f1";
}
void f2()
{
cout<<"base f2";
}
};
class d1: virtual public base
{
public:
void f1()
{
cout<<"d1's f1";
}
void f2()
{
cout<<"d1's f2";
}
};
class d2: virtual public base, public d1
{
public:
d2()
{
cout<<"d2 class construxtor\n";
}
~d2()
{
cout<<" d2 class destructor\n";
}
};
int main()
{
base* obj= new d2;
obj->f1();
obj->f2();
}
Java, Basic, who cares - it's all a bunch of tree-hugging hippy cr*p
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OK, you caught me. 
That'll learn me not to trust VS2005 
Graham
Librarians rule, Ook!
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Graham Shanks wrote: VS2005
That compiled in VS2005? I guess I can understand why, I think. Maybe.
Also yet another example why it's good to keep clear of tricksy inheritance features in C++. a) the programmer isn't 100% sure what should happen (I'm definitely including myself in that group), and b) neither's the compiler.
Java, Basic, who cares - it's all a bunch of tree-hugging hippy cr*p
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Stuart Dootson wrote: That compiled in VS2005?
No. Doesn't compile in VS 2005, 2008 or VC++ 6.0.
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The code doesn't compile with VC++ 6.0, VS 2005 and VS 2008.
(clrscr() is also undeclared in Visual C++; I vaguely recall it being in Turbo C. Perhaps he has an early version of Turbo C++ and it's not showing this as an error.)
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all right I got it...but if I am making f1() and f2() virtual in base class and d1 class, how vptr will work?
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As base is declared virtual, there is one instance of base in d2. This means (IIUC) that the functions in d1 will hide the functions in base from the point-of-view of d2 because d1 is more derived than base.
Java, Basic, who cares - it's all a bunch of tree-hugging hippy cr*p
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For the working of vptr, allow me to redirect you to one of my articles Polymorphism in C[^]
«_Superman_»
I love work. It gives me something to do between weekends.
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How to capture the currently playing song using visual c++?
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just can't understand how to make it.especially how to get the pathname of the document using begindrag message.
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Have a look at this article[^]. The list control needs to be a drop target. Look at the OnDragOver and OnDrop handlers - they mention a data object (of type COleDatObject). That object will contain data of a number of formats. I'd suggest you implement the drop target thing and then experiment with dropping explorer files onto the list control and see what data formats the data object contains, to see which is the format you want to access. CFSTR_SHELLIDLIST is probably a good one - it returns an array of PIDLISTs, one for each explorer item. You can get paths from that using SHGetPathFromIDList[^]
Java, Basic, who cares - it's all a bunch of tree-hugging hippy cr*p
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WOW!!sound so difficult!!~~
i didn't find it in mfc class wizard messages
that's too professional to me,thanks anyway
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