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The code:
class testClass {
public:
testClass() { m_buff = new BYTE[20];
memcpy(m_buff, "0123456789", 10);
};
~testClass() { delete m_buff; };
operator LPBYTE () { return m_buff; };
void show() {
cout << "m_buff = " << (__int64) (void *) m_buff << "\n\t";
for (int i = 0; i < 10; i++)
cout << m_buff[i];
cout << "\n";
};
private:
LPBYTE m_buff;
};
int main(int argc, char * argv[])
{
testClass * foo = new testClass();
foo->show();
cout << "foo = " << (__int64) (void *) foo << "\n";
cout << "(LPBYTE) foo = " << (__int64) (void *) ( (LPBYTE) foo ) << "\n\t";
for (int i = 0; i < 10; i++)
cout << ((LPBYTE) foo)[i];
cout << "\n";
delete foo;
return true;
};
And some output:
m_buff = 9180016
0123456789
foo = 9179968
(LPBYTE) foo = 9179968
-garbage-
I was expecting to see:
m_buff = 9180016
0123456789
foo = 9179968
(LPBYTE) foo = 9180016
0123456789
What am I missing ?
I know I could accomplish something similar with a member function returning the address of m_buff, but in my ever more ridiculous quest to learn more aspects of C++, I wanted to try this operator overloading bit....
As always, any guidance you can offer is greatly appreciated.
MZRmodified on Wednesday, March 17, 2010 5:11 AMP.S.
My apologies for using __int64... I know it's Microsoft specific, but I use what I've got.
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Mike the Red wrote: operator LPBYTE () { return m_buff; };
Don't you mean (LPBYTE)*foo since foo is a pointer?
/Moak
PS: You might want to add copy constructor and assignment operator to your class. Law of the Big Two, classes that handle memory/resources usually need a copy constructor and an assignment operator.
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I didn't understand your post until Pallini gave a longer explanation of this.
Thanks for your help!!!
PS:
Thanks for the suggestion about the Big Two, but I don't think so in this case.
The idea behind the class is for it to always behave as a dynamically-allocated LPBYTE, but somtimes point to a MMF, when the buffer is larger than a certain threshold.
In both cases (true LPBYTE or LPBYTE pointing to an MMF), the standard memory functions (memcpy, memset, etc.) are appropriate.
(Which is why I want the cast to point to the same place as the member pointer m_buff.)
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Good teamwork with Pallini 
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Mike the Red wrote: I was expecting to see:
Mike the Red wrote: m_buff = 9180016 0123456789 foo = 9179968 (LPBYTE) foo = 9180016 // same address as m_buff
0123456789
Mike the Red wrote: What am I missing ?
The address of the class and the one of the m_buff variable (i.e. &m_buff) are the same.
On the other hand, the address of the class differs from that of the memory buffer (i.e. m_buff).
To verify, add the following line to the show method:
cout << "&m_buff = " << (__int64) (void *) &m_buff << "\n\t";
(that happens because of dynamic memory allocation).
If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong.
-- Iain Clarke
[My articles]
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I changed the operator function to:
operator LPBYTE () { cout << "Inside operator.\n"; return &this->m_buff[0]; };
...but Inside operator. never appears in the output.
If I explicitly use foo->operator LPBYTE() then Inside operator. appears in the output but not if I use an explicit cast (LPBYTE) foo.
Any ideas on that one ?
Thanks,
MZR
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Sorry, I misunderstood the original question (I missed the cast operator overloading...  ).
You overloaded correctly it, but you should apply it to an instance of the class not to a pointer to, modify from
for (int i = 0; i < 10; i++)
cout << ((LPBYTE) foo)[i];
to
for (int i = 0; i < 10; i++)
cout << ((LPBYTE) (*foo))[i];
If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong.
-- Iain Clarke
[My articles]
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You are welcome.
If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong.
-- Iain Clarke
[My articles]
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u expect to see
(LPBYTE) foo = 9180016 and its true.. remeber m_buff is a ptr, so just check its address (using &m_buff). Hope u will see it is same as (LPBYTE)foo Means m_buff is just a ptr which holds an address somewhere in heap.
And while printing (LPBYTE)foo it provides garbage, because both foo & m_buff is dynamically allocated, so we can never say where they r in heap. To become more clear make m_buff an array, say m_buff[10], then (LPBYTE)foo will print 0123.. as u expected. Also see (void*)&m_buff = (void*)foo Hope now things are clear to u..
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I'm clear that m_buff is a pointer, and what I want is for (LPBYTE) foo to point to the same place m_buff points, not to point to m_buff itself.
My class uses the buffer size to sometimes have m_buff point to a dynamically allocated BYTE array and sometimes point to a MMF, but to allow the given buffer to be accessed as an LPBYTE regardless of the buffer 'implementation.'
It seems, though, that (LPBYTE) foo gives a pointer to the testClass object, rather than the memory pointed to by m_buff.
I think I said it in the original post, but I could accomplish this easily with a member function like
LPBYTE getBuff() { return &this->m_buff[0]; };
..but I thought this was a good opportunity to learn about operator overloading.
Thanks!
MZR
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As Moak said, try :
cout << foo->LPBYTE()[0]; virtual void BeHappy() = 0;
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This:
LPBYTE bar = foo->operator LPBYTE();
for (int i = 0; i < 10; i++)
cout << bar[i];
cout << "\n";
Gives the expected output
0123456789
....it's like my overloaded operator isn't being called when I try to explicitly cast to (LPBYTE).
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Very good
And now, as Moak said, try :
LPBYTE lpOut = *foo;
cout << lpOut[0]; virtual void BeHappy() = 0;
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hi,
i have a char array in which i m getting path of current module.i want to convert it into string before loading it.hw can i do that?
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You actually asked this question 10 minutes ago in C# forum.
If you want to convert char array to string in C++ you could use:
1) The constructor of std::string - here[^].
2) The assignment operator of std::string - hete[^].
Example:
const char* szChars = "ABCD";
std::string strFromChars(szChars);
std::string strFromChars;
strFromChars = szChars;
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char arr[MAX_PATH];
std::string str = arr; OR
wchar_t arr[MAX_PATH];
std::wstring str = arr; OR
TCHAR arr[MAX_PATH];
CString str = arr;
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Which type of string you mean.?
CString or std::string?
for CString---
CString strPath(szPathName);
for std::string--
const char* szPathName;
std::string strPath;
...
szPathName = "c:\\Test";
strPath = szPathName;
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thanx all and nuri ismail i mistakenly posted in c# forum i m doing it in vc++
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The given link works fine for me.
Maybe you should try again and if you have no success I'll send you the downloaded code. 
Regards
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Nope, still not working
I can download C# and Java examples, but C++ lib that I need is "silent" 
Must be than some odd firewall rule, or something by my side. I though that maybe link was wrong...
If it's not too much bother, please send me the lib!
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I've just sent the lib to your mail.
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Thank you very much, you are the man!!!
Best regards,
Kresimir.
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You're welcome!
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