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zakkas2483 wrote: somefun(lptSub);
MessageBox(NULL,lptSub,L"ABC",1);
Is it displaying the right string? Can you show me the "somefun" code?
He never answers anyone who replies to him. I've taken to calling him a retard, which is not fair to retards everywhere.-Christian Graus
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Thank you everybody. i did one mistake in my code while copying the in to lptMessage.
And sorry for taking your valuable time.
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zakkas2483 wrote: _tcscat( lptMessage, L"TEMP::" );
_tcscat( lptMessage, lptSub );
Why didn't you change it to
_tcscat( lptMessage, 256, L"TEMP::" );
_tcscat( lptMessage, 256, lptSub );
as I've suggested?
Sorry, I've confused _tcscat, with the secure cousin, as Hey Joe pointed out. Sorry again. 
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]
modified on Friday, March 27, 2009 1:17 PM
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Because your code makes absolutely no sense. It won't even compile. Furthermore, you are mixing the "_t..." prefix with an L"", instead of a _T() macro. Perhaps you meant to use wcscat_s?
Anyone who thinks he has a better idea of what's good for people than people do is a swine.
- P.J. O'Rourke
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My code? No sense?  I challenge you to duel, Sir. 
BTW: You're right on mixing _t with L on the other hand the code makes perfect sense.
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 parry your thrust and counter with: since when does strcat() have three parameters?
Anyone who thinks he has a better idea of what's good for people than people do is a swine.
- P.J. O'Rourke
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you're right indeed, my bad, I've confused it with the secure version . I'm sorry for the OP.
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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Can someone please tell me how to get the following to display the right binary display and not have the '1' show up at the end.
eg value (3 * 5) = 15 -> 1111
value (1 + 1) = 2 -> 1011 How do i stop it at 10.
If i change the 4 to 2 it works. If the calculations go back to 15. I don't get the 1111 only 11. What am I missing?
thanks a million
char string[64];
int value;
for (int bit = 0; bit < 4; bit++)
{
int mask = value >> bit;
if ((value & mask) == mask)
string[bit] = '1';
else
string[bit] = '0';
}
string[4] = '\0';
thanks a million
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Why not just use _itoa or _ltoa with a radix of 2?
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Yes that works. when the calculation results in a smaller number.
I change my int to _int64 and i get a 0 in the output obviously with some compile warnings about loss of data.
Is there an identical function for int64.
thanks a million.
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What about two calls to itoa?
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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FISH786 wrote: What am I missing?
Well, you need to know what the different operations do and how to use them when building your algorithm. This is a good exercise.
Some hints:
You are e.g. right shifting value by bit steps and put the result in mask. Unless bit equals zero, value and mask will most likely be different. Then you check if mask and value are the same.... 
It also looks like you're starting from the right (with the least significant bit) in value, but starts from the left in the string....
"It's supposed to be hard, otherwise anybody could do it!" - selfquote
"High speed never compensates for wrong direction!" - unknown
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Yes that was the idea. To make this work and look up how the operators and shifting works.
Thanks for your help.
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do you mean this
int bit;
for ( bit = 0; bit < 4; bit++)
{
string[3-bit] = '0' + (value & 1);
value >>= 1;
}
string[bit] = '\0';
?
However the above code generates 0010 when value=2.
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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CPallini wrote: However the above code generates 0010 when value=2
What's wrong with that?
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It is not wrong, of course...  
Anyway he may prefer 10 instead of 0010.
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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Hi FISH786,
Change following
int mask = value >> bit;
with
int mask = 8 >> bit;
This code change is applicable for 4 digit binary number only.
regards,
Divyang Mithaiwala
Software Engineer
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Your mask calculations aren't quite right (in that you aren't necessarily generating a mask with a single bit set). Using a loop, I'd implement your conversion to binary like this:
char string[64] = {0};
int value = 2;
for (int bit = 3; bit >= 0; bit--)
{
const int mask = 1 << bit;
string[(3-bit)] = (mask==(value&mask))?'1':'0';
}
string[4] = '\0';
std::cout << string << std::endl;
Java, Basic, who cares - it's all a bunch of tree-hugging hippy cr*p
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One interpretation looks like...
FISH786 wrote: char string[64];
Change this to:
char string[64] = { '\0' };FISH786 wrote: int mask = value >> bit;
Remove this.
FISH786 wrote: if ((value & mask) == mask)
Change this to:
if (value & 1) Add this below the if/else:
value = value >> 1; FISH786 wrote: string[4] = '\0';
Remove this.
When the loop ends, your binary string will be in reverse order. A simple call to strrev() will fix that.
"Old age is like a bank account. You withdraw later in life what you have deposited along the way." - Unknown
"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons
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Hi,
I am loading JPG images using follwing code.
image.Load(Image);
HBITMAP hbmp = image.Detach();
if(hbmp == NULL)
{
Error = GetLastError();
}
Load function is failing and I am getting error code 33.
I am searching in google.But I am not getting any useful information.
can anyone help me.
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From WinError.h:
//
// MessageId: ERROR_LOCK_VIOLATION
//
// MessageText:
//
// The process cannot access the file because another process has locked a portion of the file.
//
#define ERROR_LOCK_VIOLATION 33L
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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In order to get the actual message from the OS you may use FormatMessage()
Something like:
DWORD dwError = ::GetLastError();
LPTSTR lpBuffer;
CString csErrorMsg;
if (::FormatMessage(FORMAT_MESSAGE_FROM_SYSTEM | FORMAT_MESSAGE_ALLOCATE_BUFFER, NULL, dwError,
MAKELANGID(LANG_NEUTRAL, SUBLANG_SYS_DEFAULT), (LPTSTR)&lpBuffer, 0, NULL) != 0)
{
csErrorMsg.Format("%s", lpBuffer);
}
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hemlat wrote: Load function is failing and I am getting error code 33.
If Load() is failing, you should be calling GetLastError() after it, not Detach().
"Old age is like a bank account. You withdraw later in life what you have deposited along the way." - Unknown
"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons
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