Re: stdout.flush

[Steven Schveighoffer via Digitalmars-d-learn]

On 5/25/16 2:09 PM, Charles Hixson via Digitalmars-d-learn wrote:

Using:
dmd --version
DMD64 D Compiler v2.071.0

on debian Linux, and importing:
importstd.stdio;

the line:
 flush();
causes:
nt.d(29): Error: undefined identifier 'flush', did you mean function
'fflush'?

This appears solved by doing stdout.flush;  (compiles, but I'm still
writing the code) but as write, writef, etc. don't require explicitly
mentioning stdout, I think that the documentation for flush should
mention that the file must be specified.  Currently it seems to imply
that all files will be flushed.


write, writef, etc. are all forwarding functions to stdout.write, 
stdout.writef, etc. It's not that it doesn't require mentioning stdout, 
it's that you have a shortcut defined.


Just like C has printf, and fprintf, we have both. But because this 
isn't C, we don't need to change the name :)


flush as a standalone function is not supported at this time. I doubt it 
will be.


-Steve

stdout.flush

[Charles Hixson via Digitalmars-d-learn]

Using:
dmd --version
DMD64 D Compiler v2.071.0

on debian Linux, and importing:
importstd.stdio;

the line:
flush();
causes:
nt.d(29): Error: undefined identifier 'flush', did you mean function 
'fflush'?


This appears solved by doing stdout.flush;  (compiles, but I'm still 
writing the code) but as write, writef, etc. don't require explicitly 
mentioning stdout, I think that the documentation for flush should 
mention that the file must be specified.  Currently it seems to imply 
that all files will be flushed.

Re: Is stdout.flush() unsafe?

[Steven Schveighoffer via Digitalmars-d-learn]

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

Is stdout.flush() unsafe?

[tsbockman via Digitalmars-d-learn]

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`?