• Timothee Cour (18/18) Aug 17 2013 Is there any way to get the enclosing function as symbol ?
• JS (2/20) Aug 17 2013 use a string mixin?
• Nicolas Sicard (19/42) Aug 18 2013 I thought this would work but it doesn't:
• Kapps (4/22) Aug 17 2013 I don't think you can pass it in to the function, but you can use

Timothee Cour <thelastmammoth gmail.com> writes:

Is there any way to get the enclosing function as symbol ?

I'd like something like that:
alternative names would be:
__function__
__context__

----
auto fun(alias caller=__function__)(){
  //caller represents fun1!double
  return ReturnType!caller.init;
}

T fun1(T)(T x){
  assert(__function__.stringof==__FUNCTION__);
  alias fun=__function__;
  assert( is(ReturnType! __function__) == T);
  return fun();
}
void main(){fun1!double();}
----

"JS" <js.mdnq gmail.com> writes:

On Sunday, 18 August 2013 at 01:52:50 UTC, Timothee Cour wrote:
 Is there any way to get the enclosing function as symbol ?

 I'd like something like that:
 alternative names would be:
 __function__
 __context__

 ----
 auto fun(alias caller=__function__)(){
   //caller represents fun1!double
   return ReturnType!caller.init;
 }

 T fun1(T)(T x){
   assert(__function__.stringof==__FUNCTION__);
   alias fun=__function__;
   assert( is(ReturnType! __function__) == T);
   return fun();
 }
 void main(){fun1!double();}
 ----

use a string mixin?

"Nicolas Sicard" <dransic gmail.com> writes:

On Sunday, 18 August 2013 at 02:50:32 UTC, JS wrote:
 On Sunday, 18 August 2013 at 01:52:50 UTC, Timothee Cour wrote:
 Is there any way to get the enclosing function as symbol ?

 I'd like something like that:
 alternative names would be:
 __function__
 __context__

 ----
 auto fun(alias caller=__function__)(){
  //caller represents fun1!double
  return ReturnType!caller.init;
 }

 T fun1(T)(T x){
  assert(__function__.stringof==__FUNCTION__);
  alias fun=__function__;
  assert( is(ReturnType! __function__) == T);
  return fun();
 }
 void main(){fun1!double();}
 ----

 use a string mixin?

I thought this would work but it doesn't:
---
void foo(T)()
{
	bar!__FUNCTION__();
}

void bar(string Caller)()
{
	mixin("alias caller = " ~ Caller ~ ";");
}

void main()
{
	foo!double();
}
---

It works if foo isn't a template, though. The problem when foo is 
a template is that foo!double.foo seems to be an illegal 
construct for the compiler...

"Kapps" <opantm2+spam gmail.com> writes:

On Sunday, 18 August 2013 at 01:52:50 UTC, Timothee Cour wrote:
 Is there any way to get the enclosing function as symbol ?

 I'd like something like that:
 alternative names would be:
 __function__
 __context__

 ----
 auto fun(alias caller=__function__)(){
   //caller represents fun1!double
   return ReturnType!caller.init;
 }

 T fun1(T)(T x){
   assert(__function__.stringof==__FUNCTION__);
   alias fun=__function__;
   assert( is(ReturnType! __function__) == T);
   return fun();
 }
 void main(){fun1!double();}
 ----


I don't think you can pass it in to the function, but you can use 
__traits(parent) on a variable within the function to get the 
function.