August 06, 2012 Function that calculates in compile time when it can | |
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I want to write a fibonacci(n) function that calculates the result. a) if n is known at compile time, use a template b) if not, use a normal function I know how to write a template version: template fib(ulong n) { static if( n < 2 ) const fib = n; else const fib = fib!(n-1) + fib!(n-2); } But how can I 'know' if n is known at compile time to make it use the other version? (which I won't post 'cause it is fairly easy). |

August 06, 2012 Re: Function that calculates in compile time when it can | |
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Posted in reply to Minas Mina | ```
On Monday, 6 August 2012 at 12:21:38 UTC, Minas Mina wrote:
> I want to write a fibonacci(n) function that calculates the
> result.
> a) if n is known at compile time, use a template
> b) if not, use a normal function
>
> I know how to write a template version:
> template fib(ulong n)
> {
> static if( n < 2 )
> const fib = n;
> else
> const fib = fib!(n-1) + fib!(n-2);
> }
>
> But how can I 'know' if n is known at compile time to make it
> use the other version? (which I won't post 'cause it is fairly
> easy).
What exactly do you try to accomplish? Why can't you use CTFE
instead of a template?
You can check for compile time with
static if(__ctfe)
``` |

August 06, 2012 Re: Function that calculates in compile time when it can | |
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Posted in reply to Tobias Pankrath | Something like this: template fib(ulong n) { static if( n < 2 ) const fib = n; else const fib = fib!(n-1) + fib!(n-2); if( n < 2) return n; return fib(n-1) + fib(n-2); } It doesn't work of course, as I am in a template and trying to "return" something. CTFE? Is that "compile time function evaluation"? If yes, it's really slow... If I try: static x = fib(40); // fib is a normal function it takes forever and makes my pc run really slowly. |

August 06, 2012 Re: Function that calculates in compile time when it can | |
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Posted in reply to Tobias Pankrath | On Monday, 6 August 2012 at 13:46:14 UTC, Tobias Pankrath wrote: > You can check for compile time with > > static if(__ctfe) No, you can't. __ctfe is a "CTFE-ed"(?) value. But you can do something like that : http://dpaste.dzfl.pl/e3f26239 Minas : I don't know if your PC is outdated, but it's weird. :/ |

August 06, 2012 Re: Function that calculates in compile time when it can | |
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Posted in reply to Tobias Pankrath | ```
On 08/06/12 15:46, Tobias Pankrath wrote:
> On Monday, 6 August 2012 at 12:21:38 UTC, Minas Mina wrote:
>> I want to write a fibonacci(n) function that calculates the result.
>> a) if n is known at compile time, use a template
>> b) if not, use a normal function
>>
>> I know how to write a template version:
>> template fib(ulong n)
>> {
>> static if( n < 2 )
>> const fib = n;
>> else
>> const fib = fib!(n-1) + fib!(n-2);
>> }
>>
>> But how can I 'know' if n is known at compile time to make it use the other version? (which I won't post 'cause it is fairly easy).
>
> What exactly do you try to accomplish? Why can't you use CTFE instead of a template?
He wants the function to be evaluated at compile time if that is possible,
but at runtime if that can't be done; and *not* have to handle both cases
in every caller. Which I can't think of any way to do right now.
An "T fib(T)(enum T n)" overload that would enable cases like this (and
bearophiles ranged-integers) to work, has been proposed in the past (only
with 'static' instead of 'enum'; the latter fits better).
IOW a "fib(42)" call would map to more or less:
enum T c = 42;
fib(c);
and inside such an overload the argument 'n' would be treated just like if it
had been defined as:
enum T n = 42;
ie it would be cfte'able and /then/ static-if and friends would be able to
handle the rest.
artur
``` |

August 06, 2012 Re: Function that calculates in compile time when it can | |
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Posted in reply to Eyyub | ```
On Monday, 6 August 2012 at 14:25:29 UTC, Eyyub wrote:
> On Monday, 6 August 2012 at 13:46:14 UTC, Tobias Pankrath wrote:
>> You can check for compile time with
>>
>> static if(__ctfe)
> No, you can't.
>
> __ctfe is a "CTFE-ed"(?) value.
>
> But you can do something like that :
> http://dpaste.dzfl.pl/e3f26239
>
> Minas : I don't know if your PC is outdated, but it's weird. :/
It's Ubuntu 12.04 running on an Intel i5.
That's what I tried to calculate:
static x = fib(40);
ulong fib(ulong n)
{
if(n < 2)
return n;
else
return fib(n-1) + fib(n-2);
}
Maybe because it has a lot of recursion?
``` |

August 06, 2012 Re: Function that calculates in compile time when it can | |
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Posted in reply to Minas Mina | On Mon, Aug 6, 2012 at 3:54 PM, Minas Mina <minas_mina1990@hotmail.co.uk> wrote: > Something like this: > > > template fib(ulong n) > { > static if( n < 2 ) > const fib = n; > else > const fib = fib!(n-1) + fib!(n-2); > > if( n < 2) > return n; > return fib(n-1) + fib(n-2); > } > > It doesn't work of course, as I am in a template and trying to "return" > something. > > CTFE? Is that "compile time function evaluation"? If yes, it's really > slow... > If I try: > static x = fib(40); // fib is a normal function > > it takes forever and makes my pc run really slowly. Well, you're using the worst possible algorithm to calculate Fibonacci (exponential time), so it's no wonder it's taking foverer :) Then, you've to know that CT calculation is far slower than runtime calculation. My experience on this is about an order of magnitude slower, and even more. On the machine I'm currently on, fib(30) is calculated instantaneously at runtime, but it takes 4-6s at CT. Fib(40) aloready takes 4-6 s at runtime, so I won't test at CT :) To come back to your question. __ctfe should be used with a standard (non-static) if. Here I implement to Fibonacci algos, one linear in n at CT, one exponential ar RT. That's just to show that a good algo at CT can run circles around a bad algo at RT. At compile-time, getting fib(100) is instantaneous, while getting only fib(40) at RT takes a few seconds on my machine. import std.conv: to; import std.stdio; long fib(size_t n) { if(__ctfe) // compile-time, linear, sustained development { long[] temp = new long[](n+1); // dynamic array during CTFE, D rox temp[0] = 1; temp[1] = 1; size_t p = 1; while (p < n) { ++p; temp[p] = temp[p-1]+temp[p-2]; } return -temp[p]; // '-' as an indication that this indeed took place at CT } else // runtime, exponential, woohoo baby! { if (n == 0 || n == 1) return 1; else return fib(n-1)+fib(n-2); } } void main() { enum f1 = fib(100); // CT pragma(msg, "At CT, fib(100) = " ~to!string(f1)); // will be < 0 as a flag auto f2 = fib(40); // RT writeln("At RT, fib(40) = ", f2); } Don't try fib(100) at runtime! |

August 06, 2012 Re: Function that calculates in compile time when it can | |
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Posted in reply to Minas Mina | ```
On Monday, 6 August 2012 at 15:18:22 UTC, Minas Mina wrote:
> That's what I tried to calculate:
> static x = fib(40);
>
> ulong fib(ulong n)
> {
> if(n < 2)
> return n;
> else
> return fib(n-1) + fib(n-2);
> }
>
> Maybe because it has a lot of recursion?
That algorithm makes O(2^n) calls to fib. I think templates get
only expanded
once for every set of parameters, so you get memoization build in
and thus it's faster in this case.
``` |

August 06, 2012 Re: Function that calculates in compile time when it can | |
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Posted in reply to Philippe Sigaud | ```
On Monday, 6 August 2012 at 15:21:38 UTC, Philippe Sigaud wrote:
> On Mon, Aug 6, 2012 at 3:54 PM, Minas Mina
> <minas_mina1990@hotmail.co.uk> wrote:
>> Something like this:
>>
>>
>> template fib(ulong n)
>> {
>> static if( n < 2 )
>> const fib = n;
>> else
>> const fib = fib!(n-1) + fib!(n-2);
>>
>> if( n < 2)
>> return n;
>> return fib(n-1) + fib(n-2);
>> }
>>
>> It doesn't work of course, as I am in a template and trying to
>> "return"
>> something.
>>
>> CTFE? Is that "compile time function evaluation"? If yes, it's
>> really
>> slow...
>
>> If I try:
>> static x = fib(40); // fib is a normal function
>>
>> it takes forever and makes my pc run really slowly.
>
> Well, you're using the worst possible algorithm to calculate
> Fibonacci
> (exponential time), so it's no wonder it's taking foverer :)
> Then, you've to know that CT calculation is far slower than
> runtime
> calculation. My experience on this is about an order of
> magnitude
> slower, and even more. On the machine I'm currently on, fib(30)
> is
> calculated instantaneously at runtime, but it takes 4-6s at CT.
> Fib(40) aloready takes 4-6 s at runtime, so I won't test at CT
> :)
>
> To come back to your question. __ctfe should be used with a
> standard
> (non-static) if.
> Here I implement to Fibonacci algos, one linear in n at CT, one
> exponential ar RT.
> That's just to show that a good algo at CT can run circles
> around a
> bad algo at RT.
> At compile-time, getting fib(100) is instantaneous, while
> getting only
> fib(40) at RT takes a few seconds on my machine.
>
> import std.conv: to;
> import std.stdio;
>
> long fib(size_t n)
> {
> if(__ctfe) // compile-time, linear, sustained development
> {
> long[] temp = new long[](n+1); // dynamic array during CTFE,
> D rox
> temp[0] = 1;
> temp[1] = 1;
> size_t p = 1;
> while (p < n)
> {
> ++p;
> temp[p] = temp[p-1]+temp[p-2];
> }
> return -temp[p]; // '-' as an indication that this indeed
> took place at CT
> }
> else // runtime, exponential, woohoo baby!
> {
> if (n == 0 || n == 1)
> return 1;
> else
> return fib(n-1)+fib(n-2);
> }
> }
>
> void main()
> {
> enum f1 = fib(100); // CT
> pragma(msg, "At CT, fib(100) = " ~to!string(f1)); // will be <
> 0 as a flag
> auto f2 = fib(40); // RT
> writeln("At RT, fib(40) = ", f2);
> }
>
> Don't try fib(100) at runtime!
Thank you for your reply!
Haha, yeah, I knew I was using the worst possible algorithm. I
``` |

August 06, 2012 Re: Function that calculates in compile time when it can | |
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Posted in reply to Philippe Sigaud | ```
On Monday, 6 August 2012 at 15:21:38 UTC, Philippe Sigaud wrote:
> On Mon, Aug 6, 2012 at 3:54 PM, Minas Mina
> <minas_mina1990@hotmail.co.uk> wrote:
>> Something like this:
>>
>>
>> template fib(ulong n)
>> {
>> static if( n < 2 )
>> const fib = n;
>> else
>> const fib = fib!(n-1) + fib!(n-2);
>>
>> if( n < 2)
>> return n;
>> return fib(n-1) + fib(n-2);
>> }
>>
>> It doesn't work of course, as I am in a template and trying to
>> "return"
>> something.
>>
>> CTFE? Is that "compile time function evaluation"? If yes, it's
>> really
>> slow...
>
>> If I try:
>> static x = fib(40); // fib is a normal function
>>
>> it takes forever and makes my pc run really slowly.
>
> Well, you're using the worst possible algorithm to calculate
> Fibonacci
> (exponential time), so it's no wonder it's taking foverer :)
> Then, you've to know that CT calculation is far slower than
> runtime
> calculation. My experience on this is about an order of
> magnitude
> slower, and even more. On the machine I'm currently on, fib(30)
> is
> calculated instantaneously at runtime, but it takes 4-6s at CT.
> Fib(40) aloready takes 4-6 s at runtime, so I won't test at CT
> :)
>
> To come back to your question. __ctfe should be used with a
> standard
> (non-static) if.
> Here I implement to Fibonacci algos, one linear in n at CT, one
> exponential ar RT.
> That's just to show that a good algo at CT can run circles
> around a
> bad algo at RT.
> At compile-time, getting fib(100) is instantaneous, while
> getting only
> fib(40) at RT takes a few seconds on my machine.
>
> import std.conv: to;
> import std.stdio;
>
> long fib(size_t n)
> {
> if(__ctfe) // compile-time, linear, sustained development
> {
> long[] temp = new long[](n+1); // dynamic array during CTFE,
> D rox
> temp[0] = 1;
> temp[1] = 1;
> size_t p = 1;
> while (p < n)
> {
> ++p;
> temp[p] = temp[p-1]+temp[p-2];
> }
> return -temp[p]; // '-' as an indication that this indeed
> took place at CT
> }
> else // runtime, exponential, woohoo baby!
> {
> if (n == 0 || n == 1)
> return 1;
> else
> return fib(n-1)+fib(n-2);
> }
> }
>
> void main()
> {
> enum f1 = fib(100); // CT
> pragma(msg, "At CT, fib(100) = " ~to!string(f1)); // will be <
> 0 as a flag
> auto f2 = fib(40); // RT
> writeln("At RT, fib(40) = ", f2);
> }
>
> Don't try fib(100) at runtime!
Thank you for your reply!
Haha, yeah, I knew I was using the worst possible algorithm. It
was just for testing... I'm never going to use ctfe with
algorithms of that complexity again!
Ok, so I can use if(__ctfe) to define different behaviour(or not)
at compile time than at runtime. The way I want to use it,
however, I don't have to:
import std.stdio;
void main()
{
ulong f1 = fibonacci(50); // calculated at runtime
static f2 = fibonacci(50); // calculated at compile time
writeln(f);
}
// calcuated at O(1) woohoo!
ulong fibonacci(ulong n)
{
import std.math;
double
r0 = (1 + sqrt(5.0)) / 2.0,
r1 = (1 - sqrt(5.0)) / 2.0,
a = 1.0 / sqrt(5.0),
b = -1.0 / sqrt(5.0);
// fn = a*r0 + b*r1
return cast(ulong)(a * pow(r0, cast(double)n) + b * pow(r1,
cast(double)n));
}
What I was really looking for was to do something like the way
std.math.sin works. I read somewhere that if the argument is
available at compile time, it evaluates the result at compile
time.
So, if I do:
double f = sin(0.5); // calculated at compile time or not?
If it is calculated at compile time, how do I do it for my own
functions?
If not, I guess the only way is to use static or enum like you
guys showed me.
``` |

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