C / C++ / MFC

27-Mar-09 7:17

you're right indeed, my bad, I've confused it with the secure version Blush | :O

. I'm sorry for the OP.
Smile | :)

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Help with Binary

FISH78627-Mar-09 0:11

FISH786

27-Mar-09 0:11

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

Michael Schubert27-Mar-09 0:20

27-Mar-09 0:20

Why not just use _itoa or _ltoa with a radix of 2?

FISH78627-Mar-09 0:36

FISH786

27-Mar-09 0:36

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.

Michael Schubert27-Mar-09 0:40

27-Mar-09 0:40

FISH786 wrote:
Is there an identical function for int64

Here are all the variations:
http://msdn.microsoft.com/en-us/library/yakksftt(VS.80).aspx[^]

CPallini27-Mar-09 0:43

27-Mar-09 0:43

What about two calls to itoa? Smile | :)

Roger Stoltz27-Mar-09 0:33

Roger Stoltz

27-Mar-09 0:33

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.... Suspicious | :suss:

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....

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FISH78627-Mar-09 1:31

FISH786

27-Mar-09 1:31

Yes that was the idea. To make this work and look up how the operators and shifting works.
Thanks for your help.

CPallini27-Mar-09 0:35

27-Mar-09 0:35

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.

Michael Schubert27-Mar-09 0:39

27-Mar-09 0:39

CPallini wrote:
However the above code generates 0010 when value=2

What's wrong with that?

CPallini27-Mar-09 0:45

27-Mar-09 0:45

It is not wrong, of course... Big Grin | :-D

Anyway he may prefer 10 instead of 0010.
Smile | :)

Michael Schubert27-Mar-09 0:51

27-Mar-09 0:51

Ah, I see.

Divyang Mithaiwala27-Mar-09 0:42

Divyang Mithaiwala

27-Mar-09 0:42

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

Stuart Dootson27-Mar-09 0:45

Stuart Dootson

27-Mar-09 0:45

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

Problem in loading Image,GetLastError 33

David Crow27-Mar-09 3:07

David Crow

27-Mar-09 3:07

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.

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hemlat26-Mar-09 23:01

hemlat

26-Mar-09 23:01

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.

Re: Problem in loading Image,GetLastError 33

CPallini26-Mar-09 23:15

Re: Problem in loading Image,GetLastError 33

26-Mar-09 23:15

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

Michael Schubert27-Mar-09 0:50