Does a C++ namespace alias defined inside a function definition have a block, function, file, or other scope (duration of validity)?
问题:
回答1:
It's a block duration of validity. E.g If you define a namespace alias as below, the namespace alias abc would be invalid outside {...} block.
{
namespace abc = xyz;
abc::test t; //valid
}
abc::test t; //invalid
回答2:
The scope is the declarative region in which the alias is defined.
回答3:
It would have the scope of the block in which it was defined - likely to be the same as function scope unless you declare the alias inside a block within a function.
回答4:
I'm fairly certain that a namespace alias only has scope within the block it's created in, like most other sorts of identifiers. I can't check for sure at the moment, but this page doesn't seem to go against it.
回答5:
As far as I know, it's in the scope it's declared. So, if you alias in a method, then it's valid in that method, but not in another.
回答6:
Take a look at http://en.wikibooks.org/wiki/C++_Programming/Scope/Namespaces
回答7:
It is valid for the duration of the scope in which it is introduced.
Take a look at http://en.cppreference.com/w/cpp/language/namespace_alias, I trust the explanation of cppreference, it's much more standard.