Trying to compile this:

void main() @safe
{
    import std.stdio;
    stdout.flush();
}

Fails with this message:
> Error: safe function 'main' cannot access __gshared data 'stdout'

Is this necessary? If so, what are the synchronization requirements for access to `stdout`?

On 12/29/15 4:57 AM, tsbockman wrote:
> Trying to compile this:
>
> void main() @safe
> {
>      import std.stdio;
>      stdout.flush();
> }
>
> Fails with this message:
>> Error: safe function 'main' cannot access __gshared data 'stdout'
>
> Is this necessary? If so, what are the synchronization requirements for
> access to `stdout`?

Hm... what is needed is an accessor for stdout:

void main() @safe
{
   import std.stdio;
   auto safe_stdout() @trusted { return stdout; }
   safe_stdout.flush();
}

The issue is the storage class __gshared is banned from accessing in safe code (because it is subject to races). So you have to claim to the compiler "I know this is generally not safe, but I have encapsulated it in a way to make it safe". Likely, this is what we should do for all the standard streams, not being able to access streams in safe code seems a steep restriction.

-Steve