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

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

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.