std.conv.to vs. casting

[Ali Çehreli]

On 07/04/2013 09:43 AM, Joseph Rushton Wakeling wrote:

 > On 07/04/2013 06:16 PM, Ali Çehreli wrote:
 >> I am commenting without fully understanding the context: Both size_t 
and double
 >> are 64 bit types on a 64-bit system. double.mant_dig being 53, 
converting from
 >> size_t to double loses information for many values.
 >
 > Oh, bugger.  You mean that because it needs space to store the 
exponent, it has
 > a reduced number of significant figures compared to size_t?

Exactly.

 >> import std.stdio;
 >> import std.conv;
 >>
 >> void main()
 >> {
 >>      size_t i = 0x8000_0000_0000_0001;
 >>      double d0 = i.to!double;
 >>      double d1 = cast(double)i;
 >>
 >>      writefln("%s", i);
 >>      writefln("%f", d0);
 >>      writefln("%f", d1);
 >> }
 >>
 >> Prints
 >>
 >> 9223372036854775809
 >> 9223372036854775808.000000
 >> 9223372036854775808.000000
 >
 > Don't understand why the error arises here, as the number of 
significant figures
 > is the same for all the numbers ... ?

It is about the difference between the significant number of bits. The 
size_t value above uses bits 63 and 0, a range of 64 bits. double's 
53-bit mantissa has no room for bit 0.

In contrast, the following size_t value can be represented exactly in a 
double because its representation uses only 53 bits:

import std.stdio;
import std.conv;

void main()
{
     size_t i = 0x001f_ffff_ffff_ffff;
     double d = i.to!double;

     writefln("%s", i);
     writefln("%f", d);
}

Prints:

9007199254740991
9007199254740991.000000

However, set any one of the currently-unset upper 11 bits (64 - 53 == 
11), and you get an inexact conversion.

Ali