C / C++ / MFC

problems with CTime calculation

7-Mar-06 0:31

GetHostByName()[^]

Re: IP

J51219827-Mar-06 3:22

J5121982

7-Mar-06 3:22

i got it thanks

CString CIP::GetIP()
{
CString strIp;

//Init winsock
WSADATA wsaData;
int nErrorCode = WSAStartup(MAKEWORD(1,1), &wsaData);
if (nErrorCode != 0) {
//Cannot initialize winsock
return _T("");
}

char strHostName[81];
if (gethostname(strHostName, 80)==0)
{
hostent *pHost = gethostbyname(strHostName);
if (pHost->h_addrtype == AF_INET)
{
in_addr **ppip=(in_addr**)pHost->h_addr_list;

//Enumarate all addresses
while (*ppip)
{
in_addr ip=**ppip;
strIp = CString(inet_ntoa(ip));
ppip++;
if (strIp!=_T("")) {
break;
}
}
}
}

return strIp;
}

JAYARAJ

Neil scsn7-Mar-06 0:13

Neil scsn

7-Mar-06 0:13

Hi,
Im using the CTime to calculate the date when adding number of days. The
strange thing is that when I add 58 and 59 days to the same date it returns
the same date. Here is the small code:

void DateCalc(int Days)
{
int Date=1;
int Month=9;
int Year1,Month1,Day1;

CTime Check( 2006, Month,Date,0,0,0);
Check+=CTimeSpan( Days,0,0,0);

Year1=Check.GetYear();
Month1=Check.GetMonth();
Day1=Check.GetDay();
TRACE("Year %d Month %d Date %d\n",Year1,Month1,Day1);
}

... function(...)
{
DateCalc(58); //add 58 days to 1/9/2006
DateCalc(59); //add 59 days to 1/9/2006

Both TRACE(..) functions within DateCalc(..) displays :

Year 2006 Month 10 Date 29
Year 2006 Month 10 Date 29

Anyone explain this?
Neil

Nibu babu thomas7-Mar-06 1:07

Nibu babu thomas

7-Mar-06 1:07

Just try COleDateTime instead of CTime.

Nibu thomas
Software Developer

Neil scsn7-Mar-06 3:04

Neil scsn

7-Mar-06 3:04

How do I add number of days to date? I tried the below, but didnt allow me to dd days to it.

int Month=9;
int Date=1;

COleDateTime Check( 2006, Month,Date,0,0,0);
Check+=CTimeSpan( 58,0,0,0);

Neil

David Crow7-Mar-06 3:10

David Crow

7-Mar-06 3:10

NeilDevlin wrote:
Check+=CTimeSpan( 58,0,0,0);

Check+=COleDateTimeSpan( 58,0,0,0);

"Let us be thankful for the fools. But for them the rest of us could not succeed." - Mark Twain

"There is no death, only a change of worlds." - Native American Proverb

Neil scsn7-Mar-06 3:21

Neil scsn

7-Mar-06 3:21

That worked, thanks.. Smile | :)

Neil

Graham Bradshaw7-Mar-06 2:53

Graham Bradshaw

7-Mar-06 2:53

<wild guess>

British Summer Time ends on 29 Oct. My guess is that the CTime class is doing some BST/GMT stuff behind your back. I'd check the hour value of the Check variable, to see if that's changing.

</wild guess>

Finding number of bits set in a given number

David Crow7-Mar-06 3:24

David Crow

7-Mar-06 3:24

NeilDevlin wrote:
Anyone explain this?

Go ahead and output the hour, minute, and second. Notice that the first one is 00:00:00 while the second one is 23:23:00. That BST thing is affecting you. To solve this, simply add another 0 to the CTime constructor.

"Let us be thankful for the fools. But for them the rest of us could not succeed." - Mark Twain

"There is no death, only a change of worlds." - Native American Proverb

Amar Sutar7-Mar-06 0:09

Amar Sutar

7-Mar-06 0:09

How do I find the number of bits set in a given integer number without using any while/for loop? Confused | :confused:

Re: Finding number of bits set in a given number [fixed]

toxcct7-Mar-06 0:20

7-Mar-06 0:20

myInt mi = /*...*/;  //The integer to scan
 
int iCpt = 0;   //The counter of set bits
 
//For each bits in the integer
for (int i = 0; i < sizeof(mi) * 8; i++) {
 
    //If the most right bit is set
    if ((mi & 0x1) == 0x1) {
 
        //Increasing the counter
        iCpt++;
 
    }
 
    //shifting all bits of 1 position to the right
    mi >> 1;
}
 
printf("mi contains %d '1'\n", iCpt);

ps: don't cross posts the forums. delete your question on the C++/CLI forum, as it is not a managed C++ question

-- modified at 6:51 Tuesday 7th March, 2006 (thanks stephen)

Stephen Hewitt7-Mar-06 0:48

7-Mar-06 0:48

I don't believe this code will work:

v2.0 wrote:
//For each bits in the integer
for (int i = 0; i < sizeof(mi); i++) {

Say "MyInt" is of type char. In this case sizeof(char) is 1. So the loop only loops once - But it should loop 8 times. With the following modification it will work:

for (int i = 0; i < sizeof(mi)*8; i++) {

PS: This assumes 8 bit chars - A fair assumption but not universally true on all platforms.

Steve

toxcct7-Mar-06 0:51

Re: Finding number of bits set in a given number [fixed]

7-Mar-06 0:51

yep sorry, i fixed it consequently.

thanks

David Crow7-Mar-06 3:42

David Crow

7-Mar-06 3:42

v2.0 wrote:
for (int i = 0; i < sizeof(mi) * 8; i++) {

I believe that amar specifically indicated no while or for loops.

"Let us be thankful for the fools. But for them the rest of us could not succeed." - Mark Twain

"There is no death, only a change of worlds." - Native American Proverb

Re: Finding number of bits set in a given number [fixed]

toxcct7-Mar-06 3:44

7-Mar-06 3:44

yes, badKarma already notified me about that point that i did not see firstly

Stephen Hewitt7-Mar-06 0:32

7-Mar-06 0:32

Given this:

#include <iostream>
#include <bitset>
using namespace std;
 
typedef bitset<sizeof(int)*8> IntBitSet;

You can count the bits like this:

cout << IntBitSet(42).count();

Steve

toxcct7-Mar-06 0:34

7-Mar-06 0:34

that's good, but very expensive way though

Stephen Hewitt7-Mar-06 0:52

7-Mar-06 0:52

With templates only the methods actually called are compiled. Also, in general, the methods will be inline. That said the MSVC6 STL bitset class isn't as good as it could be and it will be slower then manually written code. With a good compiler (which MSVC6 is not) and a good STL there is no reason why it can't be just as efficient.

Steve

toxcct7-Mar-06 0:56

7-Mar-06 0:56

because it expands the bits to the array, and as the minimum unit that can be addressed is the byte (not the bit), then it might slower the treatment...

i'd recommend using bitarray when it's really needed, for huge bits operations for instance (refering to bjarne stroustrup reflection on the subject)...

Stephen Hewitt7-Mar-06 3:06

7-Mar-06 3:06

v2.0 wrote:
because it expands the bits to the array, and as the minimum unit that can be addressed is the byte

That is not how bitsets work. They store the data as bits and use bit operations ("and" & "or", etc) to address them. It has an embedded class called reference that makes this seamless but it does happen. Here's some code from MSVC6's implementation (I altered the formatting):

bitset<_N>& set(size_t _P, bool _X = true)
{
     if (_N <= _P)
         _Xran();
     if (_X)
         A[_P / _Nb] |= (_Ty)1 << _P % _Nb;
    else
        _A[_P / _Nb] &= ~((_Ty)1 << _P % _Nb);
    return (*this);
}

This is the whole idea behind bitsets - To have the notational convince without space overheads you alluded to above.

Steve

toxcct7-Mar-06 3:12

7-Mar-06 3:12

and don't you think this is comsuming more cpu ?

firstly, you call a function, so unless it is inlined, there's all the call stack stuff that comes to work.
then each operation is actually a call to an operator function, and in a line like _A[_P / _Nb] &= ~((_Ty)1 << _P % _Nb), you have about 7 operators called !!!

do you see what i want to show off ?

Stephen Hewitt7-Mar-06 11:53

7-Mar-06 11:53

The functions are inline - functions which are defined and not just declared in the class definition are implicitly inline. The operators involved compile down to very simple machine code instructions, for example:
/32 : sar eax,5
%32 : and eax,0000001Fh
For a function which sets or clears a numbered bit there is no avoiding the &=, |= and << operators. As I said the MSVC6 implementation could be better:
- The constructors are not efficient.
- There is no specialization for the cases when the number of bits can fit into a machine word. If there was we could avoid the [], / and % operations in these cases in the set method.
My main point however was to point out that the bitset doesn’t expand the data out to one byte per bit as you suggested. Also with a good implementation of STL and a decent compiler (I haven’t look under the hood of MCVC7’s STL) using a bitset is free (or close to).

Steve

BadKarma7-Mar-06 12:17

BadKarma

7-Mar-06 12:17

If you realy worry about cpu cycles you should write in asm or at least in inline assembly
See my post here[^]
I agree that in most situation the bitset will prefrom more then adequate.

codito ergo sum

-- modified at 18:28 Tuesday 7th March, 2006

http://graphics.stanford.edu/~seander/bithacks.html#CountBitsSetNaive[^]:

C^hris L^osinger7-Mar-06 2:07

C^hris L^osinger

7-Mar-06 2:07

const unsigned char BitsSetTable256[] = 
{
  0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4, 
  1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 
  1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 
  2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 
  1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 
  2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 
  2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 
  3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7, 
  1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 
  2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 
  2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 
  3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7, 
  2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 
  3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7, 
  3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7, 
  4, 5, 5, 6, 5, 6, 6, 7, 5, 6, 6, 7, 6, 7, 7, 8
};

unsigned int v; // count the number of bits set in 32-bit value v
unsigned int c; // c is the total bits set in v

// Option 1:
c = BitsSetTable256[v & 0xff] + 
    BitsSetTable256[(v >> 8) & 0xff] + 
    BitsSetTable256[(v >> 16) & 0xff] + 
    BitsSetTable256[v >> 24]; 

// Option 2:
unsigned char * p = (unsigned char *) &v;
c = BitsSetTable256[p[0]] + 
    BitsSetTable256[p[1]] + 
    BitsSetTable256[p[2]] +	
    BitsSetTable256[p[3]];

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toxcct7-Mar-06 2:25