I understand and I agree, this is a much better way of doing things because with my way there wouldn't been too many changes for only a few reasons, thanks!
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------- Original Message -------
Garrett May via Std-Proposals <std-proposals_at_[hidden]> - marți, 5 aprilie 2022 la 2:19 PM a scris:


> You can do the following with concepts in C++20:
> #include <concepts>#include <iostream>
> template<std::same_as<int> T>T RandomNumber(){return 7; // for simplicity}
> template<std::same_as<float> T>T RandomNumber(){return 5.2; // for simplicity again}
> int main(){std::cout << RandomNumber<int>() << std::endl;std::cout << RandomNumber<float>() << std::endl;return 0;}
> which enforces a requirement on the template function.
> The problem with casting is that RandomNumber() would have to deducible in the first place, which it isn't as you don't know the return type ahead of time. For example:
>

> auto result = RandomNumber();
> What would the type of result be? In other languages (like Rust), this detail can be omitted until very late; but in C++, adding this could be quite difficult.If the aim is to simply be able to do (T)RandomNumber(), then I don't see any advantage doing that over doing it via RandomNumber<T>() instead.
> On Tue, 5 Apr 2022 at 12:00, Gašper Ažman via Std-Proposals <std-proposals_at_[hidden]> wrote:
>

> > You can emulate that by returning a type with two conversion operators to the types you want.
> >

> > On Tue, Apr 5, 2022 at 11:59 AM PaulIRL via Std-Proposals <std-proposals_at_[hidden]> wrote:
> >

> > > Hi, I'm currently learning C++ and at first I wondered why we can have 2 functions with the same name and different parameters but, not with the same name, same parameters and a different return type, but probably this is because the c++ compiler wouldn't know which function to call, however, I think that a good solution is forcing the user (programmer) to cast the output, and that way the compiler will know which function to call.Exampleint RandomNumber() {return 7; //for simplicity}
> > > float RandomNumber() {return 5.2 //for simplicity again}
> > > one could do
> > > std::cout<<(float)RandomNumber()<<std::endl; //prints 5.2\n
> > > howeverstd::cout<<RandomNumber()<<std::endl;
> > > would throw a compiler error
> > > this would also work as expectedstd::cout<<(int)(float)RandomNumber()<<std::endl; //prints 5\n
> > >

> > > - Paul Abrudan
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> >

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