• Adel Mamin (16/17) Jul 29 2015 Hello,
• anonymous (9/13) Jul 29 2015 There are no `auto` parameters in D. You have to make it a
• Mike Parker (11/28) Jul 29 2015 Consider what would happen if my_func is compiled in a library.

"Adel Mamin" <adel mm.st> writes:

Hello,

Why dmd cannot inference the type of 'arr' in my_func() parameter?
test.d:

import std.stdio;

void my_func(auto arr)
{
   writeln(arr);
}

void main()
{
   auto arr = new int[5];

   arr = [1, 2, 3, 4, 5];

   my_func(arr);
}

 dmd test.d

test.d(3): Error: undefined identifier arr

Adel

"anonymous" <anonymous example.com> writes:

On Wednesday, 29 July 2015 at 20:20:47 UTC, Adel Mamin wrote:
 void my_func(auto arr)
 {
   writeln(arr);
 }

There are no `auto` parameters in D. You have to make it a 
template explicitly:

----
void my_func(A)(A arr)
{
   writeln(arr);
}
----

"Mike Parker" <aldacron gmail.com> writes:

On Wednesday, 29 July 2015 at 20:20:47 UTC, Adel Mamin wrote:
 Hello,

 Why dmd cannot inference the type of 'arr' in my_func() 
 parameter?
 test.d:

 import std.stdio;

 void my_func(auto arr)
 {
   writeln(arr);
 }

 void main()
 {
   auto arr = new int[5];

   arr = [1, 2, 3, 4, 5];

   my_func(arr);
 }

 dmd test.d

 test.d(3): Error: undefined identifier arr

Consider what would happen if my_func is compiled in a library. 
How would the compiler know what type arr is if there is no code 
around that is calling it? In that case, it would be unable to 
compile the function. Furthermore, what happens if my_func is 
called with multiple types? A function has one instance and one 
instance only, so arr can only ever be one type.

With templates, the code is always available, so the compiler can 
always know what the type is inside the function. And you can 
have multiple instances of any template, so it's possible for arr 
to be different types.