Re: So why doesn't popFront return an element?
Andrej Mitrovic Wrote: > Isn't it wasteful to have to call both popFront() and front() to > simultaneously remove an element from a range and return it? I mean it's an > extra function call, right? Isn't also a waste to return something that isn't going to be used? There are times it is important to just advance and either skip it or use the value in another location. In either situation I think the overhead is minimal.
Re: So why doesn't popFront return an element?
> I'm trying to understand the design of ranges. Why does popFront only set
> the front() property to return the next element in the range? Why not
> return the element in the call to popFront right away?
>
> For example code like this (which doesn't work since popFront doesn't
> return): void main()
> {
> int[] a = [1, 2];
> auto b = a.popFront;
> assert(a == [2]);
> assert(b == 1);
> }
>
> Isn't it wasteful to have to call both popFront() and front() to
> simultaneously remove an element from a range and return it? I mean it's
> an extra function call, right?
Often, it would be wasteful to have popFront return an element, since it's
perfectly legitimate to call popFront without wanting to ever see front. If
the type of front is a struct, then it would have to be copied which could
cost more than a function call. And both front and popFront would typically be
inlined anyway (at least as long as you compile with -inline), so it's
unlikely that it really costs you a function call anyway.
Really, however, I believe that it's a matter of functionality. front and
popFront do two very different things. One of them gives you the first
element; the other removes the first element. You can call them both
separately without wanting to do the other. There are plenty of cases where
you want to call popFront multiple times without ever calling front, and there
are plenty of cases where you want to call front without calling popFront. It
_definitely_ would be a problem if front were removed in favor of having
popFront return the front element, but even if you kept front and just made
popFront return an element, it's conflating two separate operations into one
function, which often is a bad idea.
In the case of popFront, I often call popFront without wanting to look at
front immediately if at all, and in most cases where you _do_ want to look at
front every time, you're probably using a foreach loop and not calling
popFront directly anyway. So, making popFront return an element wouldn't help
you much.
So, really, there's no need to make popFront return front, it's more likely to
be _less_ efficient to do that, rather than more, and it's conflating two
separate operations.
I see little to no benefit to making popFront return anything, and it could be
detrimental for it to.
- Jonathan M Davis
Re: So why doesn't popFront return an element?
Oh, I thought the compiler could optimize calls to popFront if I'm not assigning its value. Doesn't a technique exist where if a function is called and I'm not assigning its result anywhere it should simply exit the function without returning anything? It seems like a valuable compiler optimization, if that is even possible.
Re: So why doesn't popFront return an element?
Andrej Mitrovic:
> Oh, I thought the compiler could optimize calls to popFront if I'm not
> assigning its value.
>
> Doesn't a technique exist where if a function is called and I'm not
> assigning its result anywhere it should simply exit the function
> without returning anything? It seems like a valuable compiler
> optimization, if that is even possible.
The function may have different calling points, including having its pointer
taken and passed around. So you generally need the full compiled function, that
generates the output too.
If the compiler is able to perform some kind of "whole program optimization"
(today both GCC and LLVM are able to do some of it), and the compiler is able
to infer that the result of popFront is never used in the whole program, then
it might be able to remove the dead code that generates the output.
If the compiler inlines popFront at a call point, and in this call point the
result is not used, all compilers are usually able to remove the useless
computations of the result.
Some time ago I have proposed a __used_result boolean flag, usable inside
functions. If inside a function you use this value (inside a "static if"), the
function becomes a template, and it generally gets compiled two times in the
final binary. If at a calling point the result of the function doesn't get
used, the compiler calls the template instantiation where __used_result is
false, so with the static if you are able to avoid computing the result.
So this program:
int count = 0;
int foo(int x) {
count++;
static if (__used_result) {
int result = x * x;
return result;
}
}
void main() {
foo(10);
int y = foo(20);
}
Gets rewritten by the compiler to something like:
int count = 0;
auto foo(bool use_return)(int x) {
count++;
static if (use_return) {
int result = x * x;
return result;
}
}
void main() {
foo!false(10);
int y = foo!true(20);
}
That produces:
_D4test11__T3fooVb0Z3fooFiZvcomdat
pushEAX
mov ECX,FS:__tls_array
mov EDX,[ECX]
inc dword ptr _D4test5counti[EDX]
pop EAX
ret
_D4test11__T3fooVb1Z3fooFiZicomdat
pushEAX
mov ECX,FS:__tls_array
mov EDX,[ECX]
inc dword ptr _D4test5counti[EDX]
imulEAX,EAX
pop ECX
ret
Bye,
bearophile
Re: So why doesn't popFront return an element?
> Oh, I thought the compiler could optimize calls to popFront if I'm not > assigning its value. > > Doesn't a technique exist where if a function is called and I'm not > assigning its result anywhere it should simply exit the function > without returning anything? It seems like a valuable compiler > optimization, if that is even possible. That would depend on the type. For anything other than structs, if the function is inlined, it should be able to do that (if it's not inlined, it can't, because the returning is done inside of the function call). For structs, it can probably do that if there's no postblit constructor and none of its member variables have postblit constructors (or have member variables of their own which have postblit constructors, etc.). If there's a postblit constructor, however, it would have to be pure, otherwise it could have side effects, and not doing the return could then alter the behavior of the program. However, if it's pure, it probably can. I don't know what the compiler does exactly. Regardless, in the general case, the return value can't be optimized out, because that's done in the function call, not at the call site. Inlined functions should be able to get the optimization as long as it doesn't change the behavior of the program to do so, which could require that the postblit constructor is pure if a struct is being returned, and if you're returning an expression, then that expression is still going to have to be evaluated unless the compiler can determine that it has no side effects, which again, could require any functions involved to be pure. So, don't bet on such an optimization happening. dmd might manage it in some cases, but in the most obvious cases, the cost of the return is pretty low anyway (since it's either a primitive type or a reference). - Jonathan M Davis
Re: So why doesn't popFront return an element?
On 04/14/2011 01:00 AM, Andrej Mitrovic wrote:
I'm trying to understand the design of ranges. Why does popFront only set the
front() property to return the next element in the range? Why not return the
element in the call to popFront right away?
For example code like this (which doesn't work since popFront doesn't return):
void main()
{
int[] a = [1, 2];
auto b = a.popFront;
assert(a == [2]);
assert(b == 1);
}
Isn't it wasteful to have to call both popFront() and front() to simultaneously
remove an element from a range and return it? I mean it's an extra function
call, right?
I like to have three members (even if not quite necessary, this cleanly
separates notions). Why I don't understand is why empty and front are methods,
not simple data members.
Denis
--
_
vita es estrany
spir.wikidot.com
Re: So why doesn't popFront return an element?
> I'm trying to understand the design of ranges. Why does popFront only set the
front() property to return the next element in the range? Why not return the
element in the call to popFront right away?
>
> For example code like this (which doesn't work since popFront doesn't return):
> void main()
> {
> int[] a = [1, 2];
> auto b = a.popFront;
> assert(a == [2]);
> assert(b == 1);
> }
>
> Isn't it wasteful to have to call both popFront() and front() to
> simultaneously
remove an element from a range and return it? I mean it's an extra function
call,
right?
Those ranges are intended for functional programming!
Most of the time, not returning anything means much much less work. Most ranges
you'll encounter are _lazily_ evaluated. Eg:
int[]a=[1,2,...];
auto b=map!foo(map!bar1(map!bar2(a));
auto c=chain(a,b);
auto d=zip(a,b);
auto e=zip(take(repeat(d),c.length));//note infinite range here...
//etc, you get the idea
Note that this code NEVER actually calls any of foo or bar1/bar2. (if this was
not
so, the code would enter an infinite loop at repeat(d)) It keeps not calling
them
when you call popFront(). (why would you want to do this anyways? I think it has
only few applications, one of them is the implementation of opApply or new lazy
range returning functions).
When you now do e.front(), suddenly, a lot of stuff gets computed, so that you
get
a look at the first element of the range, again you usually don't really want to
do this. It is more likely that you will call "take" or a fold/reduce on that
range.
Re: So why doesn't popFront return an element?
This leads me to another question I've always wanted to ask. A call such as: auto b=map!foo(map!bar1(map!bar2(a)); This constructs a lazy range. What I'm wondering is if there are any performance issues when constructing long chains of ranges like that, since this basically routes one function to the next, which routes to the next, etc. E.g: auto a = [1, 2]; auto b = retro(a); auto c = retro(b); auto d = retro(c); Under the surface, I assume calling d.front would mean the following calls: d.front()->c.back()->b.front()->a.back() Or another example: auto b = retro(retro(a)); Can the compiler optimize the second case and convert b.front to just do one field access?
Re: So why doesn't popFront return an element?
On Thu, 14 Apr 2011 12:57:10 -0400, Andrej Mitrovic
wrote:
This leads me to another question I've always wanted to ask. A call such
as:
auto b=map!foo(map!bar1(map!bar2(a));
This constructs a lazy range. What I'm wondering is if there are any
performance issues when constructing long chains of ranges like that,
since this basically routes one function to the next, which routes to
the next
Of course. Ranges are very dependent on inlining for their performance
benefit. Which means you are depending on the compiler inlining the code
in order to achieve good performance. The compiler doesn't always inline,
and I'm not sure how deep it can go.
Or another example:
auto b = retro(retro(a));
Can the compiler optimize the second case and convert b.front to just
do one field access?
from std.range:
auto retro(Range)(Range r)
if (isBidirectionalRange!(Unqual!Range))
{
// Check for retro(retro(r)) and just return r in that case
static if (is(typeof(retro(r.source)) == Range))
{
return r.source;
}
...
So yeah, it can be optimized somewhat.
-Steve
Re: So why doesn't popFront return an element?
> On 04/14/2011 01:00 AM, Andrej Mitrovic wrote:
> > I'm trying to understand the design of ranges. Why does popFront only set
> > the front() property to return the next element in the range? Why not
> > return the element in the call to popFront right away?
> >
> > For example code like this (which doesn't work since popFront doesn't
> > return): void main()
> > {
> >
> > int[] a = [1, 2];
> > auto b = a.popFront;
> > assert(a == [2]);
> > assert(b == 1);
> >
> > }
> >
> > Isn't it wasteful to have to call both popFront() and front() to
> > simultaneously remove an element from a range and return it? I mean it's
> > an extra function call, right?
>
> I like to have three members (even if not quite necessary, this cleanly
> separates notions). Why I don't understand is why empty and front are
> methods, not simple data members.
They're properties. They could be member variables or they could be member
functions. It's up to the implementation of a particular range as to whether
they're member variables or member functions, though they're likely to be
member functions in most cases. In most cases, there wouldn't be any member
variable corresponding to empty (more likely, it's a check against the length
of the range or does something to determine whether there's more left in the
range rather than being a member variable indicating whether the range is
empty), and even with front, there could easily be a calculation of some kind
or additional checks (e.g. to make it throw if empty is true) in front.
There is no requirement that front or empty be either functions or variables.
They're properties, so they can be whatever makes the most sense for that
particular range type. And as for arrays, they _have_ to be functions, since
they're added by Phobos.
- Jonathan M Davis
Re: So why doesn't popFront return an element?
> On Thu, 14 Apr 2011 12:57:10 -0400, Andrej Mitrovic > > wrote: > > This leads me to another question I've always wanted to ask. A call such > > as: > > > > auto b=map!foo(map!bar1(map!bar2(a)); > > > > This constructs a lazy range. What I'm wondering is if there are any > > performance issues when constructing long chains of ranges like that, > > since this basically routes one function to the next, which routes to > > the next > > Of course. Ranges are very dependent on inlining for their performance > benefit. Which means you are depending on the compiler inlining the code > in order to achieve good performance. The compiler doesn't always inline, > and I'm not sure how deep it can go. IIRC, I believe that I brought that up at one point and Walter said that it could inline infinitely deep (assuming that that it makes sense of course). However, I don't think that there's much question that the inliner could be improved. If nothing else, there are too many things in the language which currently preclude inlining, and as I understand it, the inliner could use some work on how well it does at inlining what it does inline. But I haven't studied it, so I don't know how good it ultimately is. - Jonathan M Davis
Re: So why doesn't popFront return an element?
Andrei Mitrovic:
> Can the compiler optimize the second case and convert b.front to just
do one field access?
In simple cases, obviously yes:
import std.range;
import std.stdio;
void main(){
int[] a=new int[1000];
auto b=retro(retro(a));
writeln(b.front);
}
Produces:
.text:08094B64 public _Dmain
.text:08094B64 _Dmain proc near ; CODE XREF:
_D2rt6dmain24mainUiPPaZi7runMainMFZv+15p
.text:08094B64 pushebp
.text:08094B65 mov ebp, esp
.text:08094B67 mov eax, offset _D11TypeInfo_Ai6__initZ
.text:08094B6C push3E8h
.text:08094B71 pusheax
.text:08094B72 call_d_newarrayT
.text:08094B77 add esp, 8
.text:08094B7A mov eax, offset
_D3std5stdio6stdoutS3std5stdio4File
.text:08094B7F pushdword ptr [edx]
.text:08094B81 push0Ah
.text:08094B83 call
_D3std5stdio4File14__T5writeTiTaZ5writeMFiaZv
.text:08094B88 xor eax, eax
.text:08094B8A pop ebp
.text:08094B8B retn
.text:08094B8B _Dmain endp ; sp = -8
DMD even recognizes the fact, that edx contains the pointer to the first
element of b.
I do not know about more complex cases, I think it depends on DMD's inlining
caps,
which are (naturally) somewhat poorer than gcc's at the moment.
Re: So why doesn't popFront return an element?
On Thu, 14 Apr 2011 13:36:07 -0400, Timon Gehr wrote:
Andrei Mitrovic:
Can the compiler optimize the second case and convert b.front to just
do one field access?
In simple cases, obviously yes:
import std.range;
import std.stdio;
void main(){
int[] a=new int[1000];
auto b=retro(retro(a));
writeln(b.front);
}
Produces:
.text:08094B64 public _Dmain
.text:08094B64 _Dmain proc near ; CODE XREF:
_D2rt6dmain24mainUiPPaZi7runMainMFZv+15%19p
.text:08094B64 pushebp
.text:08094B65 mov ebp, esp
.text:08094B67 mov eax, offset
_D11TypeInfo_Ai6__initZ
.text:08094B6C push3E8h
.text:08094B71 pusheax
.text:08094B72 call_d_newarrayT
.text:08094B77 add esp, 8
.text:08094B7A mov eax, offset
_D3std5stdio6stdoutS3std5stdio4File
.text:08094B7F pushdword ptr [edx]
.text:08094B81 push0Ah
.text:08094B83 call
_D3std5stdio4File14__T5writeTiTaZ5writeMFiaZv
.text:08094B88 xor eax, eax
.text:08094B8A pop ebp
.text:08094B8B retn
.text:08094B8B _Dmain endp ; sp = -8
DMD even recognizes the fact, that edx contains the pointer to the first
element of b.
I do not know about more complex cases, I think it depends on DMD's
inlining caps,
which are (naturally) somewhat poorer than gcc's at the moment.
Nah, it's simpler than that:
assert(is(typeof(b) == int[]));
which is done by retro (it detects if you're retro-ing a retro range, and
just returns the original).
-Steve
Re: So why doesn't popFront return an element?
> Nah, it's simpler than that:
>
> assert(is(typeof(b) == int[]));
>
> which is done by retro (it detects if you're retro-ing a retro range, and
> just returns the original).
>
> -Steve
Makes sense. But still, I think inlining is responsible for the following code
generating identical machine code:
import std.range;
import std.stdio;
import std.algorithm;
int id(int a){return a;}
void main(){
int[] a=new int[1000];
auto b=retro(map!id(map!id(retro(a;
writeln(b.front);
}
Correct me if I'm wrong, but I think phobos does only handle the special case
retro(retro(x)) explicitly.
-Timon
