June 24, 2012
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()?
June 24, 2012
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.

June 25, 2012
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
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