Question about scope

[Walter Bright via Digitalmars-d]

On 1/27/2017 6:36 AM, Nemanja Boric wrote:
>
> ```
> void bar(scope char** c, scope int* n) @trusted
> {
> }
>
> void foo() @safe
> {
>     char[100] buf;
>     char* c = &buf[0];
>     int n;
>
>     bar(&c, &n);  // Error: cannot take address of scope local c in @safe function foo
> }
>
> void main()
> {
>     foo();
> }
> ```
>
> This doesn't compile even with dip1000, but does compile when `buf` is moved out
> of the scope. Can somebody explain why it fails and what needs to be done with it?

(I added in the error message to your example.)

Because 'buf' is local data, taking the address of 'buf' is 'scope', meaning it 
must not escape foo(). This means that 'c' is also inferred to be 'scope'.

Taking the address of a 'scope' variable is not allowed because 'scope' is not 
transitive and a 'scope pointer to a scope value' is not representable.

However, a 'ref' to a 'scope' is allowed, so your code can be written as:

```
void bar(ref scope char* c, scope int* n) @trusted
{
}

void foo() @safe
{
     char[100] buf;
     char* c = &buf[0];
     int n;

     bar(c, &n);
}

void main()
{
     foo();
}
```

If you need to build more complex structures on the stack, and have them be 
@safe, you'll need to encapsulate things with structs and ref counting.