c++: How to convert a DWORD to CHAR*?

I'm trying to convert a DWORD (that is gauranteed to be populated with a numberic value) to a char*?  How would I go about doing this?

Thanks.
i-mate05Asked:
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grg99Commented:
do you want to convert 1234567  to  "1234567"?

or

do you want to convert 0x1828342 to a  pointer to a character at address 0x1828342?

very different things.
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i-mate05Author Commented:
I'm looking to convert 123456789 to "123456789".
0
grg99Commented:
int TheNum; char Out[1000];

TheNum = 123456789;

sprintf( Out, "%d", TheNum );
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i-mate05Author Commented:
Is there a way to get the actual size of the int so the char length is exact?
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InteractiveMindCommented:
Perhaps something like


DWORD i = 123456789 ;
int size = int(i/10)+1 ;
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choo_chuCommented:
Hi,

This is primitive and there is probably a better way to find the number of numbers in a number...heh, but you could try this:

int length = 0;
int num = 1500;

while( num > 0 )
{
    num /= 10;
    length ++;
}

// length now equals 4

Regards,
choo_chu
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grg99Commented:
Dividing by ten is a good way, if a bit slow.

an alternative way, a bit faster but less precise, would be to divide by two (by bit shifting), then multiply the resulting count by log10(2),  or is it log2(10)?  


or just take ceiling( log10(num) ) + 1 and that shoul dbe close enough.



 
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jhshuklaCommented:
what about negative numbers and zero?

int x = crap;

bool y = x<0;
if ( y )
  length = ceiling( log10(num) ) + 2;
else if ( !x )
  length = 1;
else
  length = ceiling( log10(num) ) + 1;
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rajeev_devinCommented:
>> Is there a way to get the actual size of the int so the char length is exact?
Use stringstream. You don't have to specify the size then.

#include <sstream>
#include <iostream>
...
DWORD dword = 123456789;
std::stringstream stream;
stream << dword;
std::string str = stream.str();
std::cout << str << std::endl;

If you want the char* then do this

char* cPtr = str.c_str();
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