Determine if CTFE or RT
let is(CTFE == x) mean that x is a compile time constant. CTFE(x)
converts a x to this compile time constant. Passing any compile
time constant essentially turns the variable in to a compile time
constant(effectively turns it in to a template with template
parameter)
void foo(size_t i)
{
static if (is(CTFE == i))
{
pragma(msg, CTFE(i));
} else
{
writeln(i);
}
}
which, when called with a compile time constant acts effectively
like
void foo(size_t i)()
{
pragma(msg, i);
}
so
foo(1), being a CTFE'able, triggers the pragma, while foo(i) for
volatile i triggers the writeln.
Re: Determine if CTFE or RT
On Sun, 24 Jun 2018 14:43:09 +, Mr.Bingo wrote:
> let is(CTFE == x) mean that x is a compile time constant. CTFE(x)
> converts a x to this compile time constant. Passing any compile time
> constant essentially turns the variable in to a compile time
> constant(effectively turns it in to a template with template parameter)
>
You can use __ctfe:
static if (__ctfe) {
// compile-time execution
} else {
// run-time execution
}
Re: Determine if CTFE or RT
On Sunday, 24 June 2018 at 18:21:09 UTC, rjframe wrote:
On Sun, 24 Jun 2018 14:43:09 +, Mr.Bingo wrote:
let is(CTFE == x) mean that x is a compile time constant.
CTFE(x)
converts a x to this compile time constant. Passing any
compile time
constant essentially turns the variable in to a compile time
constant(effectively turns it in to a template with template
parameter)
You can use __ctfe:
static if (__ctfe) {
// compile-time execution
} else {
// run-time execution
}
no that will not work.
it cannot be a static if.
it has to be an if.
Re: Determine if CTFE or RT
On Sunday, 24 June 2018 at 18:21:09 UTC, rjframe wrote:
On Sun, 24 Jun 2018 14:43:09 +, Mr.Bingo wrote:
let is(CTFE == x) mean that x is a compile time constant.
CTFE(x)
converts a x to this compile time constant. Passing any
compile time
constant essentially turns the variable in to a compile time
constant(effectively turns it in to a template with template
parameter)
You can use __ctfe:
static if (__ctfe) {
// compile-time execution
} else {
// run-time execution
}
This does not work:
import std.stdio;
auto foo(int i)
{
if (__ctfe)
{
return 1;
} else {
return 2;
}
}
void main()
{
writeln(foo(3));
}
should print 1 but prints 2.
Re: Determine if CTFE or RT
On Sunday, 24 June 2018 at 19:10:36 UTC, Mr.Bingo wrote:
On Sunday, 24 June 2018 at 18:21:09 UTC, rjframe wrote:
On Sun, 24 Jun 2018 14:43:09 +, Mr.Bingo wrote:
let is(CTFE == x) mean that x is a compile time constant.
CTFE(x)
converts a x to this compile time constant. Passing any
compile time
constant essentially turns the variable in to a compile time
constant(effectively turns it in to a template with template
parameter)
You can use __ctfe:
static if (__ctfe) {
// compile-time execution
} else {
// run-time execution
}
This does not work:
import std.stdio;
auto foo(int i)
{
if (__ctfe)
{
return 1;
} else {
return 2;
}
}
void main()
{
writeln(foo(3));
}
should print 1 but prints 2.
you have to call foo with ctfe
enum n = foo(3);
writeln(n);
Re: Determine if CTFE or RT
On Sunday, 24 June 2018 at 20:03:19 UTC, arturg wrote:
On Sunday, 24 June 2018 at 19:10:36 UTC, Mr.Bingo wrote:
On Sunday, 24 June 2018 at 18:21:09 UTC, rjframe wrote:
On Sun, 24 Jun 2018 14:43:09 +, Mr.Bingo wrote:
let is(CTFE == x) mean that x is a compile time constant.
CTFE(x)
converts a x to this compile time constant. Passing any
compile time
constant essentially turns the variable in to a compile time
constant(effectively turns it in to a template with template
parameter)
You can use __ctfe:
static if (__ctfe) {
// compile-time execution
} else {
// run-time execution
}
This does not work:
import std.stdio;
auto foo(int i)
{
if (__ctfe)
{
return 1;
} else {
return 2;
}
}
void main()
{
writeln(foo(3));
}
should print 1 but prints 2.
you have to call foo with ctfe
enum n = foo(3);
writeln(n);
This defeats the whole purpose. The whole point is for the
compiler to automatically compute foo(3) since it can be
computed. Now, with your code, there is no way to simplify code
like
foo(3) + foo(8);
auto foo(int i)
{
if (__ctfe && i == 3)
{
return 1;
} else {
return 2;
}
}
Now, this would precompute foo(3), unbeknownst to the caller of
foo but then this requires, using your method, to write the code
like
enum x = foo(3);
x + foo(8);
We would have to know which values foo would return as compile
time values and what not.
foo(x) + foo(y)
could not work simultaneously for both compile time and run time
variables.
e.g.,
enum x = 4;
enum y = 4;
foo(x) + foo(y)
would not precompute the values even though x and y are enum's,
we have to do
enum x = 4;
enum y = 4;
enum xx = foo(x);
enum yy = foo(y);
xx + yy;
and if we changed x or y to a run time variable we'd then have to
rewrite the expressions since your technique will then fail.
int x = 4;
enum y = 4;
enum xx = foo(x); // Invalid
enum yy = foo(y);
xx + yy;
The compiler should be easily able to figure out that foo(3) can
be precomputed(ctfe'ed) and do so. It can already do this, as you
say, by forcing enum on it. Why can't the compiler figure it out
directly?
The problem here, if you didn't understand, is that one can't get
foo to be "precomputed" IF it can be done, but IF NOT then it
will just get the full computation. Because one does not
necessarily know when and where and how foo will be
precomputed(or even have completely different behavior for ctfe
vs rtfe), one can't use two different methods that have the same
syntax.
Re: Determine if CTFE or RT
On Monday, June 25, 2018 05:03:26 Mr.Bingo via Digitalmars-d-learn wrote: > The compiler should be easily able to figure out that foo(3) can > be precomputed(ctfe'ed) and do so. It can already do this, as you > say, by forcing enum on it. Why can't the compiler figure it out > directly? The big problem with that is that determining whether the calculation can be done at compile time or not means solving the halting problem. In general, the only feasible way to do it would be to attempt it for every function call and then give up when something didn't work during CTFE, which would balloon compilation times and likely cause the compiler to run out of memory on a regular basis given how it currently manages memory and how CTFE tends to use a lot of memory. It was decided ages ago that the best approach to the problem was to use CTFE only when CTFE was actually required. If an expression is used in a context where its value must be known at compile time, then it's evaluated at compile time; otherwise, it isn't. The compiler never attempts CTFE as an optimization or because it thinks that it might be possible to evaluate the value at compile time. As things stand, it should be pretty trivial to be able to look at an expression and determine whether it's evaluated at compile time or not based on how it's used. If you're looking to have the compiler figure out when to do CTFE based on the fact that an expression could theoretically be evaluated at compile time, or because you want the compiler to optimize using CTFE, then you're going to be disappointed, because that's never how CTFE has worked, and I'd be _very_ surprised if it ever worked any differently. - Jonathan M Davis
Re: Determine if CTFE or RT
On Monday, 25 June 2018 at 05:14:31 UTC, Jonathan M Davis wrote:
On Monday, June 25, 2018 05:03:26 Mr.Bingo via
Digitalmars-d-learn wrote:
The compiler should be easily able to figure out that foo(3)
can be precomputed(ctfe'ed) and do so. It can already do this,
as you say, by forcing enum on it. Why can't the compiler
figure it out directly?
The big problem with that is that determining whether the
calculation can be done at compile time or not means solving
the halting problem. In general, the only feasible way to do it
would be to attempt it for every function call and then give up
when something didn't work during CTFE, which would balloon
compilation times and likely cause the compiler to run out of
memory on a regular basis given how it currently manages memory
and how CTFE tends to use a lot of memory.
It was decided ages ago that the best approach to the problem
was to use CTFE only when CTFE was actually required. If an
expression is used in a context where its value must be known
at compile time, then it's evaluated at compile time;
otherwise, it isn't. The compiler never attempts CTFE as an
optimization or because it thinks that it might be possible to
evaluate the value at compile time. As things stand, it should
be pretty trivial to be able to look at an expression and
determine whether it's evaluated at compile time or not based
on how it's used.
If you're looking to have the compiler figure out when to do
CTFE based on the fact that an expression could theoretically
be evaluated at compile time, or because you want the compiler
to optimize using CTFE, then you're going to be disappointed,
because that's never how CTFE has worked, and I'd be _very_
surprised if it ever worked any differently.
- Jonathan M Davis
The docs say that CTFE is used only when explicit, I was under
the impression that it would attempt to optimize functions if
they could be computed at compile time. The halting problem has
nothing to do with this. The ctfe engine already complains when
one recurses to deep, it is not difficult to have a time out
function that cancels the computation within some user definable
time limit... and since fail can simply fall through and use the
rtfe, it is not a big deal.
The problem then, if D can't arbitrarily use ctfe, means that
there should be a way to force ctfe optionally!
This means that the compiler will force ctfe if the input values
are known, just like normal but if they are not known then it
just treats the call as non-ctfe.
so, instead of
enum x = foo(y); // invalid or valid depending on if y is known
at CT
we have
cast(enum)foo(y)
which, hypothetically of course, would attempt to precompute
foo(y) as in the first case but if it is not precomputable then
just calls it at compile time.
So, the above code becomes
static if (ctfeable(y))
enum x = foo(y);
return x; // compile time version ("precomputed")
else
return foo(y) // runtime version
The semantic above is defined for something like cast(enum)(might
be confusing but anything could be used that works better).
hence
auto x = cast(enum)foo(y);
will result in x being an enum if y is an enum(since chaining can
then occur), else a runtime variable with the return of foo
calculated at runtime.
So, this has nothing to do with the halting problem. I'm not
asking for the compiler to do the impossible, I'm asking for a
notation that combines to different syntaxes in to a composite
pattern so one can benefit from the other. Don't make it harder
than it seems, it's a pretty easy concept.
It might even be possible to do this in a template:
import std.stdio;
auto IsCTFE()
{
if (__ctfe) return true;
return false;
}
template AutoEnum(alias s)
{
static if (IsCTFE)
{
pragma(msg, "CTFE!");
alias AutoEnum = s;
}
else
{
pragma(msg, "RTFE!");
alias AutoEnum = s;
}
}
double foo(int x) { return 3.42322*x; }
void main()
{
int x = 3;
writeln(AutoEnum!(foo(x)));
}
So, the problem is that if we replace x with 3 the above code is
executed at compile time(a precomputation).
But the way it is it won't allow it to be computed even at
runtime! There is no reason that the compiler can't just replace
the code with `writeln(foo(x));` for RTFE... yet it errors
*RATHER THAN* default to RTFE!
The simple fix is to allow for the alternative to not try to ctfe
and just compute the value at runtime.
Re: Determine if CTFE or RT
On Monday, June 25, 2018 05:47:30 Mr.Bingo via Digitalmars-d-learn wrote:
> The problem then, if D can't arbitrarily use ctfe, means that
> there should be a way to force ctfe optionally!
If you want to use CTFE, then give an enum the value of the expression you
want calculated. If you want to do it in place, then use a template such as
template ctfe(alias exp)
{
enum ctfe = exp;
}
so that you get stuff like func(ctfe!(foo(42))). I would be extremely
surprised if the compiler is ever changed to just try CTFE just in case it
will work as an optimization. That would make it harder for the programmer
to understand what's going on, and it would balloon compilation times. If
you want to write up a DIP on the topic and argue for rules on how CTFE
could and should function with the compiler deciding to try CTFE on some
basis rather than it only being done when it must be done, then you're free
to do so.
https://github.com/dlang/DIPs
But I expect that you will be sorely disappointed if you ever expect the
compiler to start doing CTFE as an optimization. It's trivial to trigger it
explicitly on your own, and compilation time is valued far too much to waste
it on attempting CTFE when in the vast majority of cases, it's going to
fail. And it's worked quite well thus far to have it work only cases when
it's actually needed - especially with how easy it is to make arbitrary code
run during CTFE simply by doing something like using an enum.
- Jonathan M Davis
Re: Determine if CTFE or RT
On Monday, 25 June 2018 at 07:02:24 UTC, Jonathan M Davis wrote:
On Monday, June 25, 2018 05:47:30 Mr.Bingo via
Digitalmars-d-learn wrote:
The problem then, if D can't arbitrarily use ctfe, means that
there should be a way to force ctfe optionally!
If you want to use CTFE, then give an enum the value of the
expression you want calculated. If you want to do it in place,
then use a template such as
template ctfe(alias exp)
{
enum ctfe = exp;
}
so that you get stuff like func(ctfe!(foo(42))). I would be
extremely surprised if the compiler is ever changed to just try
CTFE just in case it will work as an optimization. That would
make it harder for the programmer to understand what's going
on, and it would balloon compilation times. If you want to
write up a DIP on the topic and argue for rules on how CTFE
could and should function with the compiler deciding to try
CTFE on some basis rather than it only being done when it must
be done, then you're free to do so.
https://github.com/dlang/DIPs
But I expect that you will be sorely disappointed if you ever
expect the compiler to start doing CTFE as an optimization.
It's trivial to trigger it explicitly on your own, and
compilation time is valued far too much to waste it on
attempting CTFE when in the vast majority of cases, it's going
to fail. And it's worked quite well thus far to have it work
only cases when it's actually needed - especially with how easy
it is to make arbitrary code run during CTFE simply by doing
something like using an enum.
- Jonathan M Davis
You still don't get it!
It is not trivial! It is impossible to trigger it! You are
focused far too much on the optimization side when it is only an
application that takes advantage of the ability for rtfe to
become ctfe when told, if it is possible.
I don't know how to make this any simpler, sorry... I guess we'll
end it here.
Re: Determine if CTFE or RT
On Monday, 25 June 2018 at 08:05:53 UTC, Mr.Bingo wrote:
On Monday, 25 June 2018 at 07:02:24 UTC, Jonathan M Davis wrote:
On Monday, June 25, 2018 05:47:30 Mr.Bingo via
Digitalmars-d-learn wrote:
The problem then, if D can't arbitrarily use ctfe, means that
there should be a way to force ctfe optionally!
If you want to use CTFE, then give an enum the value of the
expression you want calculated. If you want to do it in place,
then use a template such as
template ctfe(alias exp)
{
enum ctfe = exp;
}
so that you get stuff like func(ctfe!(foo(42))). I would be
extremely surprised if the compiler is ever changed to just
try CTFE just in case it will work as an optimization. That
would make it harder for the programmer to understand what's
going on, and it would balloon compilation times. If you want
to write up a DIP on the topic and argue for rules on how CTFE
could and should function with the compiler deciding to try
CTFE on some basis rather than it only being done when it must
be done, then you're free to do so.
https://github.com/dlang/DIPs
But I expect that you will be sorely disappointed if you ever
expect the compiler to start doing CTFE as an optimization.
It's trivial to trigger it explicitly on your own, and
compilation time is valued far too much to waste it on
attempting CTFE when in the vast majority of cases, it's going
to fail. And it's worked quite well thus far to have it work
only cases when it's actually needed - especially with how
easy it is to make arbitrary code run during CTFE simply by
doing something like using an enum.
- Jonathan M Davis
You still don't get it!
It is not trivial! It is impossible to trigger it! You are
focused far too much on the optimization side when it is only
an application that takes advantage of the ability for rtfe to
become ctfe when told, if it is possible.
I don't know how to make this any simpler, sorry... I guess
we'll end it here.
I am not sure that I understood it right, but there is a way to
detect the status of a parameter:
My question was different, but I wished to get a ctRegex! or
regex used depending on the expression:
import std.regex:replaceAll,ctRegex,regex;
auto reg(alias var)(){
static if (__traits(compiles, {enum ctfeFmt = var;}) ){
// "Promotion" to compile time value
enum ctfeReg = var ;
pragma(msg, "ctRegex used");
return(ctRegex!ctfeReg);
}else{
return(regex(var));
pragma(msg,"regex used");
}
}
}
So now I can always use reg!("") and let the compiler decide.
To speed up compilation I made an additional switch, that when
using DMD (for development)
alway the runtime version is used.
The trick is to use the alias var in the declaration and check if
it can be assigned to enum.
The only thing is now, that you now always use the !() compile
time parameter to call the function. Even, when in the end is
translated to an runtime call.
reg!("") and not reg("...").
Re: Determine if CTFE or RT
On 06/25/2018 07:47 AM, Mr.Bingo wrote:
The docs say that CTFE is used only when explicit, I was under the
impression that it would attempt to optimize functions if they could be
computed at compile time. The halting problem has nothing to do with
this. The ctfe engine already complains when one recurses to deep, it is
not difficult to have a time out function that cancels the computation
within some user definable time limit... and since fail can simply fall
through and use the rtfe, it is not a big deal.
The problem then, if D can't arbitrarily use ctfe, means that there
should be a way to force ctfe optionally!
A D compiler is free to precompute whatever it sees fit, as an
optimization. It's just not called "CTFE" then, and `__ctfe` will be
false during that kind of precomputation.
For example, let's try compiling this code based on an earlier example
of yours:
int main()
{
return foo(3) + foo(8);
}
int foo(int i)
{
return __ctfe && i == 3 ? 1 : 2;
}
`dmd -O -inline` compiles that to:
<_Dmain>:
0: 55 push rbp
1: 48 8b ecmovrbp,rsp
4: b8 04 00 00 00 moveax,0x4
9: 5d poprbp
a: c3 ret
As expected, `ldc2 -O` is even smarter:
<_Dmain>:
0: b8 04 00 00 00 moveax,0x4
5: c3 ret
Both compilers manage to eliminate the calls to `foo`. They have been
precomputed. `__ctfe` was false, though, because the term "CTFE" only
covers the forced/guaranteed kind of precomputation, not the optimization.
Re: Determine if CTFE or RT
On Monday, 25 June 2018 at 09:36:45 UTC, Martin Tschierschke
wrote:
I am not sure that I understood it right, but there is a way to
detect the status of a parameter:
My question was different, but I wished to get a ctRegex! or
regex used depending on the expression:
import std.regex:replaceAll,ctRegex,regex;
auto reg(alias var)(){
static if (__traits(compiles, {enum ctfeFmt = var;}) ){
// "Promotion" to compile time value
enum ctfeReg = var ;
pragma(msg, "ctRegex used");
return(ctRegex!ctfeReg);
}else{
return(regex(var));
pragma(msg,"regex used");
}
}
}
So now I can always use reg!("") and let the compiler
decide.
To speed up compilation I made an additional switch, that when
using DMD (for development)
alway the runtime version is used.
The trick is to use the alias var in the declaration and check
if it can be assigned to enum.
The only thing is now, that you now always use the !() compile
time parameter to call the function. Even, when in the end is
translated to an runtime call.
reg!("") and not reg("...").
Now try reg!("prefix" ~ var) or reg!(func(var)). This works in
some limited cases, but falls apart when you try something more
involved. It can sorta be coerced into working by passing lambdas:
template ctfe(T...) if (T.length == 1) {
import std.traits : isCallable;
static if (isCallable!(T[0])) {
static if (is(typeof({enum a = T[0]();}))) {
enum ctfe = T[0]();
} else {
alias ctfe = T[0];
}
} else {
static if (is(typeof({enum a = T[0];}))) {
enum ctfe = T[0];
} else {
alias ctfe = T[0];
}
}
}
string fun(string s) { return s; }
unittest {
auto a = ctfe!"a";
string b = "a";
auto c = ctfe!"b";
auto d = ctfe!("a" ~ b); // Error: variable b cannot be read
at compile time
auto e = ctfe!(() => "a" ~ b);
auto f = ctfe!(fun(b)); // Error: variable b cannot be read
at compile time
auto g = ctfe!(() => fun(b));
}
--
Simen
Re: Determine if CTFE or RT
On Monday, 25 June 2018 at 10:49:26 UTC, Simen Kjærås wrote:
On Monday, 25 June 2018 at 09:36:45 UTC, Martin Tschierschke
wrote:
I am not sure that I understood it right, but there is a way
to detect the status of a parameter:
My question was different, but I wished to get a ctRegex! or
regex used depending on the expression:
import std.regex:replaceAll,ctRegex,regex;
auto reg(alias var)(){
static if (__traits(compiles, {enum ctfeFmt = var;}) ){
// "Promotion" to compile time value
enum ctfeReg = var ;
pragma(msg, "ctRegex used");
return(ctRegex!ctfeReg);
}else{
return(regex(var));
pragma(msg,"regex used");
}
}
}
So now I can always use reg!("") and let the compiler
decide.
To speed up compilation I made an additional switch, that when
using DMD (for development)
alway the runtime version is used.
The trick is to use the alias var in the declaration and check
if it can be assigned to enum.
The only thing is now, that you now always use the !() compile
time parameter to call the function. Even, when in the end is
translated to an runtime call.
reg!("") and not reg("...").
Now try reg!("prefix" ~ var) or reg!(func(var)). This works in
some limited cases, but falls apart when you try something more
involved. It can sorta be coerced into working by passing
lambdas:
template ctfe(T...) if (T.length == 1) {
import std.traits : isCallable;
static if (isCallable!(T[0])) {
static if (is(typeof({enum a = T[0]();}))) {
enum ctfe = T[0]();
} else {
alias ctfe = T[0];
}
} else {
static if (is(typeof({enum a = T[0];}))) {
enum ctfe = T[0];
} else {
alias ctfe = T[0];
}
}
}
string fun(string s) { return s; }
unittest {
auto a = ctfe!"a";
string b = "a";
auto c = ctfe!"b";
auto d = ctfe!("a" ~ b); // Error: variable b cannot be
read at compile time
auto e = ctfe!(() => "a" ~ b);
auto f = ctfe!(fun(b)); // Error: variable b cannot be read
at compile time
auto g = ctfe!(() => fun(b));
}
--
Simen
This doesn't work, the delegate only hides the error until you
call it.
auto also does not detect enum. Ideally it should be a manifest
constant if precomputed... this allows chaining of optimizations.
auto x = 3;
auto y = foo(x);
the compiler realizes x is an enum int and then it can also
precompute foo(x).
Since it converts to a runtime type immediately it prevents any
optimizations and template tricks.
