Let's say I have an interface:
interface IPerson
{
int Age { get; set; }
string Name { get; set; }
bool Alive { get; set; }
}
and a Class:
public class Person : IPerson
{
public int Age { get; set; }
public string Name { get; set; }
}
That would not compile since Person does not implement the Alive Property.
What I would like to know is if there is a way to have the same behaviour, if Person adds an extra property that is not found in its interface.
interface IPerson
{
int Age { get; set; }
string Name { get; set; }
}
and a Class:
public class Person : IPerson
{
public int Age { get; set; }
public string Name { get; set; }
public bool Alive { get; set; } <---- This should prevent it from compiling as well.
}
I would want it to not compile as well, or at the very least give me a compile warning.
No. Interfaces define what members an object must implement. They cannot define members that an object can't implement. You could potentially use your own custom, or third party code analysis tools, to identify cases like this, but there is nothing in the language itself that would support it.
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