Re: mixin template question

[lobo via Digitalmars-d-learn]

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