Re: integral to floating point conversion
On Sunday, 3 July 2016 at 13:41:27 UTC, Ola Fosheim Grøstad wrote: On Sunday, 3 July 2016 at 11:49:15 UTC, Andrei Alexandrescu wrote: Well to be more precise here's what I'm looking for. When you compare an integral with a floating point number, the integral is first converted to floating point format. I.e. for long x and double y, x == y is the same as double(x) == y. Now, say we want to eliminate the "bad" cases of this comparison. Those would make it non-transitive. Consider two distinct longs x1 and x2. If they convert to the same double y, then x1 == y and x2 == y are true, which is contradictory with x1 != x2. If you assume round-to-even rounding mode then you get unique representations for integers in the ranges as other people have suggested: int_to_float : [-((1<<24)-1) , (1<<24)-1] int_to_double : [-((1<<53)-1) , (1<<53)-1] Seems to me there are some assumptions being made here that only the permanent storage bits are in play. You've covered the implicit leading bit, but what about possible guard digits, depending on where the values are currently located in the hardware? Not that I've thought it through, but it's something to think about.
Re: integral to floating point conversion
On Sunday, 3 July 2016 at 11:49:15 UTC, Andrei Alexandrescu wrote: Well to be more precise here's what I'm looking for. When you compare an integral with a floating point number, the integral is first converted to floating point format. I.e. for long x and double y, x == y is the same as double(x) == y. Now, say we want to eliminate the "bad" cases of this comparison. Those would make it non-transitive. Consider two distinct longs x1 and x2. If they convert to the same double y, then x1 == y and x2 == y are true, which is contradictory with x1 != x2. If you assume round-to-even rounding mode then you get unique representations for integers in the ranges as other people have suggested: int_to_float : [-((1<<24)-1) , (1<<24)-1] int_to_double : [-((1<<53)-1) , (1<<53)-1]
Re: integral to floating point conversion
On 07/03/2016 05:52 AM, Guillaume Boucher wrote: On Sunday, 3 July 2016 at 09:08:14 UTC, Ola Fosheim Grøstad wrote: On Saturday, 2 July 2016 at 20:17:59 UTC, Andrei Alexandrescu wrote: So what's the fastest way to figure that an integral is convertible to a floating point value precisely (i.e. no other integral converts to the same floating point value)? Thanks! -- Andrei If it is within what the mantissa can represent then it is easy. But you also have to consider the cases where the mantissa is shifted. So the real answer is: n is an unsigned 64 bit integer mbits = representation bits for mantissa +1 tz = trailing_zero_bits(n) lz = leading_zero_bits(n) assert(mbits >= (64 - tz - lz)) This is the correct answer for another definition of "precisely convertible", not the one Andrei gave. Well to be more precise here's what I'm looking for. When you compare an integral with a floating point number, the integral is first converted to floating point format. I.e. for long x and double y, x == y is the same as double(x) == y. Now, say we want to eliminate the "bad" cases of this comparison. Those would make it non-transitive. Consider two distinct longs x1 and x2. If they convert to the same double y, then x1 == y and x2 == y are true, which is contradictory with x1 != x2. Andrei
Re: integral to floating point conversion
On Sunday, 3 July 2016 at 09:52:38 UTC, Guillaume Boucher wrote: This is the correct answer for another definition of "precisely convertible", not the one Andrei gave. True, I see now that he actually asked for unique representation, not precisely convertible.
Re: integral to floating point conversion
On Sunday, 3 July 2016 at 09:08:14 UTC, Ola Fosheim Grøstad wrote: On Saturday, 2 July 2016 at 20:17:59 UTC, Andrei Alexandrescu wrote: So what's the fastest way to figure that an integral is convertible to a floating point value precisely (i.e. no other integral converts to the same floating point value)? Thanks! -- Andrei If it is within what the mantissa can represent then it is easy. But you also have to consider the cases where the mantissa is shifted. So the real answer is: n is an unsigned 64 bit integer mbits = representation bits for mantissa +1 tz = trailing_zero_bits(n) lz = leading_zero_bits(n) assert(mbits >= (64 - tz - lz)) This is the correct answer for another definition of "precisely convertible", not the one Andrei gave.
Re: integral to floating point conversion
On Saturday, 2 July 2016 at 20:17:59 UTC, Andrei Alexandrescu wrote: So what's the fastest way to figure that an integral is convertible to a floating point value precisely (i.e. no other integral converts to the same floating point value)? Thanks! -- Andrei If it is within what the mantissa can represent then it is easy. But you also have to consider the cases where the mantissa is shifted. So the real answer is: n is an unsigned 64 bit integer mbits = representation bits for mantissa +1 tz = trailing_zero_bits(n) lz = leading_zero_bits(n) assert(mbits >= (64 - tz - lz))
Re: integral to floating point conversion
On Saturday, 2 July 2016 at 20:30:03 UTC, Walter Bright wrote: On 7/2/2016 1:17 PM, Andrei Alexandrescu wrote: So what's the fastest way to figure that an integral is convertible to a floating point value precisely (i.e. no other integral converts to the same floating point value)? Thanks! -- Andrei Test that its absolute value is <= the largest unsigned value represented by the float's mantissa bits. That has to be '<' given the condition that no other integer converts to the same value. Although 2^n can be represented exactly, 2^n+1 would be converted to the same float value.
Re: integral to floating point conversion
On Saturday, 2 July 2016 at 20:49:27 UTC, Andrei Alexandrescu wrote: On 7/2/16 4:30 PM, Walter Bright wrote: On 7/2/2016 1:17 PM, Andrei Alexandrescu wrote: So what's the fastest way to figure that an integral is convertible to a floating point value precisely (i.e. no other integral converts to the same floating point value)? Thanks! -- Andrei Test that its absolute value is <= the largest unsigned value represented by the float's mantissa bits. What is the largest unsigned value represented by the float's mantissa bits? I recall there is a constant somewhere for it. -- Andrei 2uL^^float.mant_dig And the same for double. For real (on x86), mant_dig is 64, so it can represent any ulong precisely. -- Simen
Re: integral to floating point conversion
On Saturday, 2 July 2016 at 20:17:59 UTC, Andrei Alexandrescu
wrote:
So what's the fastest way to figure that an integral is
convertible to a floating point value precisely (i.e. no other
integral converts to the same floating point value)? Thanks! --
Andrei
bool isConvertible(T) (long n) if (is(T == float) || is(T ==
double)) {
pragma(inline, true);
static if (is(T == float)) {
return (((n+(n>>63))^(n>>63))&0xff00UL) == 0;
} else {
return (((n+(n>>63))^(n>>63))&0xffe0UL) == 0;
}
}
Re: integral to floating point conversion
On 7/2/2016 1:49 PM, Andrei Alexandrescu wrote: On 7/2/16 4:30 PM, Walter Bright wrote: On 7/2/2016 1:17 PM, Andrei Alexandrescu wrote: So what's the fastest way to figure that an integral is convertible to a floating point value precisely (i.e. no other integral converts to the same floating point value)? Thanks! -- Andrei Test that its absolute value is <= the largest unsigned value represented by the float's mantissa bits. What is the largest unsigned value represented by the float's mantissa bits? I recall there is a constant somewhere for it. -- Andrei https://en.wikipedia.org/wiki/Single-precision_floating-point_format 24 bits. So it can store +-0x0FFF_
Re: integral to floating point conversion
On Saturday, 2 July 2016 at 20:49:27 UTC, Andrei Alexandrescu wrote: On 7/2/16 4:30 PM, Walter Bright wrote: On 7/2/2016 1:17 PM, Andrei Alexandrescu wrote: So what's the fastest way to figure that an integral is convertible to a floating point value precisely (i.e. no other integral converts to the same floating point value)? Thanks! -- Andrei Test that its absolute value is <= the largest unsigned value represented by the float's mantissa bits. What is the largest unsigned value represented by the float's mantissa bits? I recall there is a constant somewhere for it. -- Andrei Isn't it 2^24-1?
Re: integral to floating point conversion
On 7/2/16 4:30 PM, Walter Bright wrote: On 7/2/2016 1:17 PM, Andrei Alexandrescu wrote: So what's the fastest way to figure that an integral is convertible to a floating point value precisely (i.e. no other integral converts to the same floating point value)? Thanks! -- Andrei Test that its absolute value is <= the largest unsigned value represented by the float's mantissa bits. What is the largest unsigned value represented by the float's mantissa bits? I recall there is a constant somewhere for it. -- Andrei
Re: integral to floating point conversion
On 7/2/2016 1:17 PM, Andrei Alexandrescu wrote: So what's the fastest way to figure that an integral is convertible to a floating point value precisely (i.e. no other integral converts to the same floating point value)? Thanks! -- Andrei Test that its absolute value is <= the largest unsigned value represented by the float's mantissa bits.
integral to floating point conversion
So what's the fastest way to figure that an integral is convertible to a floating point value precisely (i.e. no other integral converts to the same floating point value)? Thanks! -- Andrei
