• mastrost (28/28) Dec 26 2008 Hello,
• Walter Bright (3/4) Dec 26 2008 The 'this' reference must also be considered an argument to a pure
• mastrost (2/7) Dec 27 2008 Ok thank you, I understand better now.

mastrost <titi.mastro free.fr> writes:

Hello,

I have searched on the D2 documentation pages and on this forum, but I did not
really found any documentation about pure member functions, so please excuse me
if a make you repeat.

I was very surprised to see this code compile (dmd 2.022):

class A{
    private:
        int x;
    public:
        pure int f() {
            return x;
        }
        int g() {
            ++x;
            return x;
        }
}

void main(){

    A a=new A;
    writefln(a.f()); //prints 0
    a.g();
    writefln(a.f()); //prints 1
}

- So what kind of function can be considered as a pure member function ?

- In my exemple, is A.f really considered as pure by the compiler, because I
explicitly added the keyword 'pure' ?

- Is a D compiler a good gift for christmas?

- Is this assertion true:
    "If we wanted to have true purity for member functions, it would mean that
as soon as a member function is pure, all other functions in the same module
would have to be pure, expect the constructors"

- (equivalent to the previous question) If inside a module only the
constructors of the classes are not pure, can I be sure that no behaviour like
what we see in my example will happen (the first call to a.f() gives different
result from the second call to a.f() ) ?

Thank you all

Walter Bright <newshound1 digitalmars.com> writes:

mastrost wrote:
 I was very surprised to see this code compile (dmd 2.022):

The 'this' reference must also be considered an argument to a pure 
function: a.f() is f(a) as far as purity checking goes.

mastrost <titi.mastro free.fr> writes:

Walter Bright Wrote:

 mastrost wrote:
 I was very surprised to see this code compile (dmd 2.022):

 
 The 'this' reference must also be considered an argument to a pure 
 function: a.f() is f(a) as far as purity checking goes.

Ok thank you, I understand better now.