@safe function with __gshared as default parameter value

Anonymouse zorael at gmail.com
Wed Apr 8 16:53:05 UTC 2020

import std.stdio;


__gshared int gshared = 42;

void foo(int i = gshared)

void main()

This currently works; `foo` is `@safe` and prints the value of 
`gshared`. Changing the call in main to `foo(gshared)` errors.

Should it work, and can I expect it to keep working?

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