On 3/21/2014 11:23 AM, Steven Schveighoffer wrote:
On Fri, 21 Mar 2014 14:18:05 -0400, Walter Bright <newshou...@digitalmars.com>
wrote:
On 3/21/2014 12:59 AM, monarch_dodra wrote:
Ok. That's a fair point. So in that case, our function is pointing at "data",
and is allowed to mutate it, and observe its state.
Now, if *another* piece of code is doing the same thing at the same time
(potentially mutating "data", does that still violate purity?
A mutex essentially reads and writes a global flag, which other functions can
also read and write.
That makes it NOT pure.
No, that's not the case. A mutex does not write a global flag, it writes a
shared flag. And the flag is passed into it.
Here's a litmus test you can use for purity. Consider:
int foo() pure;
{
auto a = foo();
}
Now, since 'a' is never used, we can delete the assignment, and since foo is
pure, we can delete foo():
{
}
Is this program distinguishable from the former? If not, then foo() is not pure.
