Re: mixin template question

[Paul D Anderson via Digitalmars-d-learn]

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.

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