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