• tsbockman (9/11) Dec 29 2015 Trying to compile this:
• Steven Schveighoffer (15/25) Dec 29 2015 Hm... what is needed is an accessor for stdout:

tsbockman <thomas.bockman gmail.com> writes:

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`?

Steven Schveighoffer <schveiguy yahoo.com> writes:

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