On Sunday, 28 June 2015 at 09:19:16 UTC, Tofu Ninja wrote:
module main;
import std.stdio;
void main(string[] args)
{
auto d = foo();
writeln(d()); // prints 25
}
auto foo()
{
int x = 4;
pure int delegate() d = delegate()
{
return x*x;
};
writeln(d()); // prints 16
x = 5;
writeln(d()); // prints 25
return d;
}
I can see that after foo returns, then d will truly be pure as
x will no longer be modifiable, but just before that, it seems
d is not actually pure. Is this the intended behavior?
Here's a similar example where it's a bit clearer what's going on:
----
struct S
{
int x = 0;
int d() pure {return ++this.x;}
}
void main()
{
S s;
int delegate() pure d = &s.d;
import std.stdio;
writeln(d()); /* prints "1" */
writeln(d()); /* prints "2" */
writeln(d()); /* prints "3" */
}
----
`d` isn't all that pure, right? You're seeing the concept of
"weak purity" in action. In D, the attribute `pure` really just
means "doesn't access global mutable state". The function is
still allowed to mutate any arguments it gets, including hidden
ones. Here the method `d` has a hidden `this` parameter, and it
can mutate data through it. With a nested function, the enclosing
context is passed via a hidden parameter.
Properly/mathematically/strongly pure functions have additional
requirements on top of being marked `pure`. Read more here:
http://klickverbot.at/blog/2012/05/purity-in-d/
