• Paul D Anderson (18/18) Apr 11 2015 I don't understand why the following code compiles and runs
• lobo (26/44) Apr 11 2015 As the manual says (snippet below) the surrounding scope
• Paul D Anderson (2/55) Apr 11 2015 Thanks.

"Paul D Anderson" <claude.reins msnmail.com> writes:

I don't understand why the following code compiles and runs 
without an error:

import std.stdio;

mixin template ABC(){
   int abc() { return 3; }
}

mixin ABC;

int abc() { return 4; }

void main()
{
   writefln("abc() = %s", abc());
}

Doesn't the mixin ABC create a function with the same signature 
as the "actual function" abc()?

It compiles with both included and writes "abc() = 4". If I 
comment out the actual function then it writes "abc() = 3". The 
actual function takes precedence, but why don't they conflict?

Paul

"lobo" <swamplobo gmail.com> writes:

On Sunday, 12 April 2015 at 03:51:03 UTC, Paul D Anderson wrote:
 I don't understand why the following code compiles and runs 
 without an error:

 import std.stdio;

 mixin template ABC(){
   int abc() { return 3; }
 }

 mixin ABC;

 int abc() { return 4; }

 void main()
 {
   writefln("abc() = %s", abc());
 }

 Doesn't the mixin ABC create a function with the same signature 
 as the "actual function" abc()?

 It compiles with both included and writes "abc() = 4". If I 
 comment out the actual function then it writes "abc() = 3". The 
 actual function takes precedence, but why don't they conflict?

 Paul

As the manual says (snippet below) the surrounding scope 
overrides mixin

http://dlang.org/template-mixin.html

---
Mixin Scope
The declarations in a mixin are ‘imported’ into the surrounding 
scope. If the name of a declaration in a mixin is the same as a 
declaration in the surrounding scope, the surrounding declaration 
overrides the mixin one:
int x = 3;

mixin template Foo()
{
     int x = 5;
     int y = 5;
}

mixin Foo;
int y = 3;

void test()
{
     writefln("x = %d", x);  // prints 3
     writefln("y = %d", y);  // prints 3
}
---

bye,
lobo

"Paul D Anderson" <claude.reins msnmail.com> writes:

On Sunday, 12 April 2015 at 04:04:43 UTC, lobo wrote:
 On Sunday, 12 April 2015 at 03:51:03 UTC, Paul D Anderson wrote:
 I don't understand why the following code compiles and runs 
 without an error:

 import std.stdio;

 mixin template ABC(){
  int abc() { return 3; }
 }

 mixin ABC;

 int abc() { return 4; }

 void main()
 {
  writefln("abc() = %s", abc());
 }

 Doesn't the mixin ABC create a function with the same 
 signature as the "actual function" abc()?

 It compiles with both included and writes "abc() = 4". If I 
 comment out the actual function then it writes "abc() = 3". 
 The actual function takes precedence, but why don't they 
 conflict?

 Paul

 As the manual says (snippet below) the surrounding scope 
 overrides mixin

 http://dlang.org/template-mixin.html

 ---
 Mixin Scope
 The declarations in a mixin are ‘imported’ into the surrounding 
 scope. If the name of a declaration in a mixin is the same as a 
 declaration in the surrounding scope, the surrounding 
 declaration overrides the mixin one:
 int x = 3;

 mixin template Foo()
 {
     int x = 5;
     int y = 5;
 }

 mixin Foo;
 int y = 3;

 void test()
 {
     writefln("x = %d", x);  // prints 3
     writefln("y = %d", y);  // prints 3
 }
 ---

 bye,
 lobo

Thanks.