• simendsjo <simendsjo gmail.com> Jun 24 2012
• "Tobias Pankrath" <tobias pankrath.net> Jun 24 2012
• "Kenji Hara" <k.hara.pg gmail.com> Jun 25 2012

simendsjo <simendsjo gmail.com> writes:

import std.exception;
import std.traits;

struct Ranged(T, T min, T max) {
     T _value = min;
     typeof(this) opAssign(V : T)(V value) {
         enforce(value >= min);
         enforce(value <= max);
         _value = value;
         return this;
     }
     alias _value this;
}

void f(int i) {
     i = 1000;
}

void g(T)(T i) if(isIntegral!T){
     i = 1000;
}

void main() {
     Ranged!(int, 10, 20) v;
     v = 10; // ok
     v = 20; // ok
     f(v); // auch
     g(v); // ok, exception
}

Is there a way to ensure the struct is used in f() without using templates  
as in g()?

"Tobias Pankrath" <tobias pankrath.net> writes:

On Sunday, 24 June 2012 at 13:16:53 UTC, simendsjo wrote:
 import std.exception;
 import std.traits;

 struct Ranged(T, T min, T max) {
     T _value = min;
     typeof(this) opAssign(V : T)(V value) {
         enforce(value >= min);
         enforce(value <= max);
         _value = value;
         return this;
     }
     alias _value this;
 }

 void f(int i) {
     i = 1000;
 }

 void g(T)(T i) if(isIntegral!T){
     i = 1000;
 }

 void main() {
     Ranged!(int, 10, 20) v;
     v = 10; // ok
     v = 20; // ok
     f(v); // auch
     g(v); // ok, exception
 }

 Is there a way to ensure the struct is used in f() without 
 using templates as in g()?


This should currently be impossible. The compiler would need to 
automatically promote f to a template function.

"Kenji Hara" <k.hara.pg gmail.com> writes:

On Sunday, 24 June 2012 at 13:16:53 UTC, simendsjo wrote:
 import std.exception;
 import std.traits;

 struct Ranged(T, T min, T max) {
     T _value = min;
     typeof(this) opAssign(V : T)(V value) {
         enforce(value >= min);
         enforce(value <= max);
         _value = value;
         return this;
     }
     alias _value this;
 }

 void f(int i) {
     i = 1000;
 }

 void g(T)(T i) if(isIntegral!T){
     i = 1000;
 }

 void main() {
     Ranged!(int, 10, 20) v;
     v = 10; // ok
     v = 20; // ok
     f(v); // auch
     g(v); // ok, exception
 }

 Is there a way to ensure the struct is used in f() without 
 using templates as in g()?


The type of 'alias this' symbol works as like the super class of 
user defined class.
Therefore, Ranged!(...) is always implicitly convertible to int 
by 'alias _value this', and there is no way to disable such 
conversion.

Kenji Hara