• bearophile <bearophileHUGS lycos.com> Sep 19 2010
• Don <nospam nospam.com> Sep 20 2010
• "Steven Schveighoffer" <schveiguy yahoo.com> Sep 20 2010
• Don <nospam nospam.com> Sep 20 2010
• "Steven Schveighoffer" <schveiguy yahoo.com> Sep 21 2010

bearophile <bearophileHUGS lycos.com> writes:

Jonathan M Davis:

 I assume that if you declare a member function as pure, then all of its 
 parameters - including the invisible this - are included in that. That is, if 
 all of them - including the invisible this - have the same value, then the 
 result will be the same.


This D2 program runs with no errors, and here there isn't a D language/compiler
bug:

struct Foo {
    int x;
    this (int xx) { this.x = xx; }
    pure int bar() { return x; }
}
void main() {
    Foo f = Foo(1);
    assert(f.bar() == 1);
    f.x *= 2;
    assert(f.bar() == 2);
}

Bye,
bearophile

Don <nospam nospam.com> writes:

bearophile wrote:
 Jonathan M Davis:
 
 I assume that if you declare a member function as pure, then all of its 
 parameters - including the invisible this - are included in that. That is, if 
 all of them - including the invisible this - have the same value, then the 
 result will be the same.


 This D2 program runs with no errors, and here there isn't a D
language/compiler bug:
 
 struct Foo {
     int x;
     this (int xx) { this.x = xx; }
     pure int bar() { return x; }
 }
 void main() {
     Foo f = Foo(1);
     assert(f.bar() == 1);
     f.x *= 2;
     assert(f.bar() == 2);
 }
 
 Bye,
 bearophile


You do need to be careful about concluding how 'pure' works based on the 
current behaviour of the compiler.
There's a trap here. What if you use a hypothetical startTimer() 
function which executes a delegate every few clock ticks?

void main() {
     Foo f = Foo(1);
     startTimer( () { f.x++; });
     scope(exit)
	killTimer();

     assert(f.bar() == 1); // may fail!
     f.x *= 2;
     assert(f.bar() == 2);
}

I actually think that 'pure' on a member function can only mean, it's 
cacheably pure if and only if 'this' can be cast to immutable. Which 
includes the important case where the call is made from a pure function 
(this implies that the 'this' pointer is either a local variable of a 
pure function, or an immutable object).
Since pure functions cannot call impure functions, they can't do any of 
this nasty asynchronous stuff.

In fact I think in general, that's how pure should work: once you're 
inside 'pure', you should be able to pass even mutable objects into pure 
functions. This is possible because although it's mutable, the entire 
code that modifies it is confined to a single function. So the compiler 
can determine if it is cacheably pure without performing any kind of 
whole program analysis.

But this is just my opinion. I don't know if it will eventually work 
that way.

"Steven Schveighoffer" <schveiguy yahoo.com> writes:

On Mon, 20 Sep 2010 15:45:10 -0400, Don <nospam nospam.com> wrote:

 bearophile wrote:
 Jonathan M Davis:

 I assume that if you declare a member function as pure, then all of  
 its parameters - including the invisible this - are included in that.  
 That is, if all of them - including the invisible this - have the same  
 value, then the result will be the same.


 language/compiler bug:
  struct Foo {
     int x;
     this (int xx) { this.x = xx; }
     pure int bar() { return x; }
 }
 void main() {
     Foo f = Foo(1);
     assert(f.bar() == 1);
     f.x *= 2;
     assert(f.bar() == 2);
 }
  Bye,
 bearophile


 You do need to be careful about concluding how 'pure' works based on the  
 current behaviour of the compiler.
 There's a trap here. What if you use a hypothetical startTimer()  
 function which executes a delegate every few clock ticks?

 void main() {
      Foo f = Foo(1);
      startTimer( () { f.x++; });
      scope(exit)
 	killTimer();

      assert(f.bar() == 1); // may fail!
      f.x *= 2;
      assert(f.bar() == 2);
 }


Wouldn't f have to be shared for this to be asynchronous?

 I actually think that 'pure' on a member function can only mean, it's  
 cacheably pure if and only if 'this' can be cast to immutable. Which  
 includes the important case where the call is made from a pure function  
 (this implies that the 'this' pointer is either a local variable of a  
 pure function, or an immutable object).
 Since pure functions cannot call impure functions, they can't do any of  
 this nasty asynchronous stuff.


I think it's ok for a function to be pure if all the arguments are  
unshared, regardless of immutability.  However, in order to cache the  
return value, the reference itself must not be used as the key, but the  
entire data of the reference.  Even if it's immutable, wouldn't you not  
want to cache the return values between two identical immutable objects?

-Steve

Don <nospam nospam.com> writes:

Steven Schveighoffer wrote:
 On Mon, 20 Sep 2010 15:45:10 -0400, Don <nospam nospam.com> wrote:
 
 bearophile wrote:
 Jonathan M Davis:

 I assume that if you declare a member function as pure, then all of 
 its parameters - including the invisible this - are included in 
 that. That is, if all of them - including the invisible this - have 
 the same value, then the result will be the same.


 language/compiler bug:
  struct Foo {
     int x;
     this (int xx) { this.x = xx; }
     pure int bar() { return x; }
 }
 void main() {
     Foo f = Foo(1);
     assert(f.bar() == 1);
     f.x *= 2;
     assert(f.bar() == 2);
 }
  Bye,
 bearophile


 You do need to be careful about concluding how 'pure' works based on 
 the current behaviour of the compiler.
 There's a trap here. What if you use a hypothetical startTimer() 
 function which executes a delegate every few clock ticks?

 void main() {
      Foo f = Foo(1);
      startTimer( () { f.x++; });
      scope(exit)
     killTimer();

      assert(f.bar() == 1); // may fail!
      f.x *= 2;
      assert(f.bar() == 2);
 }


 Wouldn't f have to be shared for this to be asynchronous?


That's an excellent point. 'pure' was put into the language long before 
'shared' and '__gshared'. It could now just mean, "doesn't use static, 
globals, shared, or __gshared".

And then cachable pure is just: pure, + all reference parameters are 
immutable.

If this becomes the rule, it seems likely that pure functions would 
become far more common than impure ones.

 I actually think that 'pure' on a member function can only mean, it's 
 cacheably pure if and only if 'this' can be cast to immutable. Which 
 includes the important case where the call is made from a pure 
 function (this implies that the 'this' pointer is either a local 
 variable of a pure function, or an immutable object).
 Since pure functions cannot call impure functions, they can't do any 
 of this nasty asynchronous stuff.


 I think it's ok for a function to be pure if all the arguments are 
 unshared, regardless of immutability.  However, in order to cache the 
 return value, the reference itself must not be used as the key, but the 
 entire data of the reference.  Even if it's immutable, wouldn't you not 
 want to cache the return values between two identical immutable objects?


Possibly, but my guess is that it would take too long to check.

"Steven Schveighoffer" <schveiguy yahoo.com> writes:

On Mon, 20 Sep 2010 16:26:44 -0400, Don <nospam nospam.com> wrote:

 Steven Schveighoffer wrote:
  I think it's ok for a function to be pure if all the arguments are  
 unshared, regardless of immutability.  However, in order to cache the  
 return value, the reference itself must not be used as the key, but the  
 entire data of the reference.  Even if it's immutable, wouldn't you not  
 want to cache the return values between two identical immutable objects?


 Possibly, but my guess is that it would take too long to check.


This is why I hate the idea of automatic caching -- how does the compiler  
know that it would be too long?  What if the operation takes 15 seconds,  
and to do the memcmp takes 15 milliseconds?

-Steve