Assume, we have two files:
FooBar.h
#pragma once
namespace external
{
namespace internal
{
struct Foo
{
int data;
};
struct Bar
{
Foo GetFoo();
}
}
}
and FooBar.cpp:
#include "FooBar.h"
namespace external
{
namespace internal
{
}
}
When generating implementation for Bar::GetFoo(), Visual Assist generates the following code in FooBar.cpp:
#include "FooBar.h"
namespace external
{
namespace internal
{
external::internal::Foo Bar::GetFoo()
{
throw std::logic_error("Method is not implemented");
}
}
}
There is no need to specify full qualified name external::internal::Foo withing namespace external::internal. Is it possible to generate just the following code
#include "FooBar.h"
namespace external
{
namespace internal
{
Foo Bar::GetFoo()
{
throw std::logic_error("Method is not implemented");
}
}
}
Of course if type Foo was located in a different namespace, it would make sence to specify fully qualified name as return type.
Am I missed something?
Omit full namespace name when generating a method
Posted - Jun 09 2017 : 5:27:48 PM
