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ConvertAll - Conversia unei liste mult mai usor

Avem urmatorul scenariu. O lista de elemente de tip A, pe care trebuie sa o convertim la o lista de tip B. Pentru a putea rezolva aceasta problema putem să implementam operatorul implicit sau explicit de conversie, iar apoi să iteram prin lista pentru a face conversia la tipul B.
List<A> listaA;
List<B> listaB;
...
foreach(A a in listaA)
{
listaB.Add(a);
}

sau
listaB.foreach(item=>listaB.Add(item));
Ambele soluții sunt rezonabile, dar e puțin mai greu de înțeles ca in spate se face o conversie. Pentru a face codul mai clar putem să facem in felul următor:
listaB = listaA.ConvertAll<B>(item => item)
Ce am scris mai sus? Fiecare element din listaA convertestel in tipul B si adaug-al in colecția care rezulta. item => item am putut sa scrie deoarece am presupus ca un operator de conversie implicita sau explicita intre cele doua tipuri a fost deja implementat.
In cazul in care acesta nu ar fi implementat am avea ceva asemanator:
Din
private B ConvertAToB(A a) { return new B(); ... }
...
List<A> listaA;
List<B> listaB;
...
foreach(A a in listaA)
{
listaB.Add(ConvertAToB(a));
}
Am avea:
private B ConvertAToB(A a) { return new B(); ... }
...
listaB = listaA.ConvertAll<B>(item => ConvertAToB(item));
Cea ce am scris mai sus se poate scrie si:
listaB = listaA.ConvertAll<B>(ConvertAToB);
Chiar daca folosim ConvertAll obtinem acelasi rezultat, doar ca folosind ConvertAll obtinem un cod mai clar. Urmatorul dezvoltator care o sa vina dupa noi o sa inteleaga ca aici facem oconversie.

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