# Convert UINT64 to double (MSVC6)

Dear Experts,

MSVC6 can not internally convert unsigned __int64 to double. However, VC.NET 2003 (among others) can do the conversion. In addition GCC for Intel Architectures can internally convert unsigned long long, also 64-bit unsigned integer, to double.

Does anyone have a good work-around for MSVC6? I'm using the code snippet shown below, but I find it unsatisfactory. First of all, the code does not use the full range, since for MSVC6 there is a conversion to signed 64-bit integer, and also the snippet is ugly and disturbs the more 'normal' use of static_cast<double>(UINT64 u64). Is it possible to make a template specialization for static_cast<double>(UINT64 u64) specifically for MSVC6? Alternatively, does anyone know the compiler intrinsic mathematics for the proper conversion of UINT64 to double so that I can at least correct my code snippet?

Thanks for your help.

Sincerely, Chris.

inline double UINT64_to_double(const UINT64& u)
{
#if defined (_MSC_VER) && (_MSC_VER <= 1200)

return static_cast<double>(static_cast<INT64>(u));

#else

return static_cast<double>(u);

#endif
}
LVL 1
Asked:
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Commented:
How many significant bits in a UINT64?    64 I'd guess.

How many significant bits in a double?   Less than 64.

What do you want to do if the UINT64 doesnt fit?

0

Commented:
This should do it on an Intel platform (i.e. assuming little-endian):

inline double UINT64_to_double(const UINT64& u)
{
#if defined (_MSC_VER) && (_MSC_VER <= 1200)
static const double multiplier = static_cast<double>(std::numeric_limits<UINT32>::max())+1.0;
const UINT32* p_low  = reinterpret_cast<const UINT32*>(&u);
const UINT32* p_high = p_low+1;
return *p_high*multiplier + *p_low;
#else
return static_cast<double>(u);
#endif
}
0

Commented:
> What do you want to do if the UINT64 doesnt fit?

It is a floating point representation. See the following:
--------8<--------
#include <iostream>
#include <limits>
#include <iomanip>

typedef __int64 UINT64;

template<class T> void info(const char *name)
{
std::cout
<< name
<< std::scientific
<< std::setprecision(3) /* Good enough for the max */
<< " data bits = " << std::numeric_limits<T>::digits
<< ", max = " << static_cast<double>(std::numeric_limits<T>::max())
<< '\n';
}

#define INFO(T) info<T>(#T)

int main()
{
INFO(UINT64);
INFO(double);
}
--------8<--------

UINT64 data bits = 63 max = 9.223e+018
double data bits = 53 max = 1.798e+308

double has a greater range because it uses an exponent.
0

Commented:
Right, it has a wider RANGE, but TEN fewer bits of precision.

So a random 64-bit integer has only one chance in 1024 of being accurately represented as a double.

0

Author Commented:
Thank you for all of these clear responses. Indeed, I had never really taken the time to clearly think about this conversion. The 53 bits of the ieee754 double can not identically represent UINT64.

In fact, this recognition has caused me to have to re-think the developmental issue to which this question pertains. The development issue involves converting result = (u64 * 1000000) / scale. Previously I had done a 'cheap' conversion using the cast to double of 1000000.0. But upon reflection, I noted that I actually want to do an exact conversion. This is no problem, but I will have to dig into the tools for extended integers.

Thanks everyone for clearing up this issue.

All of the responses were helpful and I must split the points equally. I can't figure out how to increase the points.

This issue is, in my opinion, now closed.

Sincerely, Chris.
0

Commented:
It might be worth your while looking at boost's big_int - see http://groups.yahoo.com/group/boost/files/big_integer/
0
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