Two-stage templated alias not resolvable in constructor?
0
I am trying to work with a two-stage templated alias, but I cannot get the alias to be accepted as a constructor parameter. Is there a way to do this?
An example of what I am trying to do is seen in the following code:
#include <memory>
template <typename Bar>
class Foo
{
public:
using FooPtr = std::shared_ptr<Foo<Bar> >;
static FooPtr getFooPtr(Bar someBar) { return std::make_shared<Foo<Bar> >(someBar); }
Foo(Bar bar) : _bar(bar) {}
Bar getBar() { return _bar; }
private:
Bar _bar;
};
template <typename Bar>
class Foobar
{
public:
using FB = Foo<Bar>;
Foobar(FB::FooPtr fooPtr)
: _fooPtr(fooPtr)
{
}
FB::FooPtr getFooPtr() { return _fooPtr; }
private:
FB::FooPtr _fooPtr;
};
int main()
{
Foobar myFoobar<int>(f::getFooPtr(4));
return myFoobar.getFooPtr()->getBar();
}
When I compile this with g++, I get the following:
dev@ubuntu:~/test/fsm$ g++ -std=c++1z test.cpp
test.cpp:23:23: error: expected ‘)’ before ‘fooPtr’
Foobar(FB::FooPtr fooPtr)
^
test.cpp:28:5: error: need ‘typename’ before ‘Foobar<Bar>::FB::FooPtr’ because ‘Foobar<Bar>::FB’ is a dependent scope
FB::FooPtr getFooPtr() { return _fooPtr; }
^
test.cpp:31:5: error: need ‘typename’ before ‘Foobar<Bar>::FB::FooPtr’ because ‘Foobar<Bar>::FB’ is a dependent scope
FB::FooPtr _fooPtr;
^
test.cpp: In function ‘int main()’:
test.cpp:36:11: error: missing template arguments before ‘myFoobar’
Foobar myFoobar<int>(f::getFooPtr(4));
^
test.cpp:37:11: error: ‘myFoobar’ was not declared in this scope
return myFoobar.getFooPtr()->getBar();
Is the compiler not recognizing the alias FB in the Foobar constuctor? Or is it a problem with resolving Foo::FooPtr to a type?
Thanks!
templates alias
asked Nov 13 '18 at 3:13
Moe42
333
add a comment |
0
I am trying to work with a two-stage templated alias, but I cannot get the alias to be accepted as a constructor parameter. Is there a way to do this?
An example of what I am trying to do is seen in the following code:
#include <memory>
template <typename Bar>
class Foo
{
public:
using FooPtr = std::shared_ptr<Foo<Bar> >;
static FooPtr getFooPtr(Bar someBar) { return std::make_shared<Foo<Bar> >(someBar); }
Foo(Bar bar) : _bar(bar) {}
Bar getBar() { return _bar; }
private:
Bar _bar;
};
template <typename Bar>
class Foobar
{
public:
using FB = Foo<Bar>;
Foobar(FB::FooPtr fooPtr)
: _fooPtr(fooPtr)
{
}
FB::FooPtr getFooPtr() { return _fooPtr; }
private:
FB::FooPtr _fooPtr;
};
int main()
{
Foobar myFoobar<int>(f::getFooPtr(4));
return myFoobar.getFooPtr()->getBar();
}
When I compile this with g++, I get the following:
dev@ubuntu:~/test/fsm$ g++ -std=c++1z test.cpp
test.cpp:23:23: error: expected ‘)’ before ‘fooPtr’
Foobar(FB::FooPtr fooPtr)
^
test.cpp:28:5: error: need ‘typename’ before ‘Foobar<Bar>::FB::FooPtr’ because ‘Foobar<Bar>::FB’ is a dependent scope
FB::FooPtr getFooPtr() { return _fooPtr; }
^
test.cpp:31:5: error: need ‘typename’ before ‘Foobar<Bar>::FB::FooPtr’ because ‘Foobar<Bar>::FB’ is a dependent scope
FB::FooPtr _fooPtr;
^
test.cpp: In function ‘int main()’:
test.cpp:36:11: error: missing template arguments before ‘myFoobar’
Foobar myFoobar<int>(f::getFooPtr(4));
^
test.cpp:37:11: error: ‘myFoobar’ was not declared in this scope
return myFoobar.getFooPtr()->getBar();
Is the compiler not recognizing the alias FB in the Foobar constuctor? Or is it a problem with resolving Foo::FooPtr to a type?
Thanks!
templates alias
asked Nov 13 '18 at 3:13
Moe42
333
The error message is clear: "error: need ‘typename’ before ‘Foobar<Bar>::FB::FooPtr’". You have to addtypenamebeforeFB::FooPtr
– max66
Nov 13 '18 at 3:18
You are completely right. I missed it because I was looking at the first error message (from line 23) which does not specify that 'typename' is missing. Adding 'typename' does fix all the cases. Thanks!
– Moe42
Nov 14 '18 at 2:05
add a comment |
0
0
0
I am trying to work with a two-stage templated alias, but I cannot get the alias to be accepted as a constructor parameter. Is there a way to do this?
An example of what I am trying to do is seen in the following code:
#include <memory>
template <typename Bar>
class Foo
{
public:
using FooPtr = std::shared_ptr<Foo<Bar> >;
static FooPtr getFooPtr(Bar someBar) { return std::make_shared<Foo<Bar> >(someBar); }
Foo(Bar bar) : _bar(bar) {}
Bar getBar() { return _bar; }
private:
Bar _bar;
};
template <typename Bar>
class Foobar
{
public:
using FB = Foo<Bar>;
Foobar(FB::FooPtr fooPtr)
: _fooPtr(fooPtr)
{
}
FB::FooPtr getFooPtr() { return _fooPtr; }
private:
FB::FooPtr _fooPtr;
};
int main()
{
Foobar myFoobar<int>(f::getFooPtr(4));
return myFoobar.getFooPtr()->getBar();
}
When I compile this with g++, I get the following:
dev@ubuntu:~/test/fsm$ g++ -std=c++1z test.cpp
test.cpp:23:23: error: expected ‘)’ before ‘fooPtr’
Foobar(FB::FooPtr fooPtr)
^
test.cpp:28:5: error: need ‘typename’ before ‘Foobar<Bar>::FB::FooPtr’ because ‘Foobar<Bar>::FB’ is a dependent scope
FB::FooPtr getFooPtr() { return _fooPtr; }
^
test.cpp:31:5: error: need ‘typename’ before ‘Foobar<Bar>::FB::FooPtr’ because ‘Foobar<Bar>::FB’ is a dependent scope
FB::FooPtr _fooPtr;
^
test.cpp: In function ‘int main()’:
test.cpp:36:11: error: missing template arguments before ‘myFoobar’
Foobar myFoobar<int>(f::getFooPtr(4));
^
test.cpp:37:11: error: ‘myFoobar’ was not declared in this scope
return myFoobar.getFooPtr()->getBar();
Is the compiler not recognizing the alias FB in the Foobar constuctor? Or is it a problem with resolving Foo::FooPtr to a type?
Thanks!
templates alias
asked Nov 13 '18 at 3:13
Moe42
333
I am trying to work with a two-stage templated alias, but I cannot get the alias to be accepted as a constructor parameter. Is there a way to do this?
An example of what I am trying to do is seen in the following code:
#include <memory>
template <typename Bar>
class Foo
{
public:
using FooPtr = std::shared_ptr<Foo<Bar> >;
static FooPtr getFooPtr(Bar someBar) { return std::make_shared<Foo<Bar> >(someBar); }
Foo(Bar bar) : _bar(bar) {}
Bar getBar() { return _bar; }
private:
Bar _bar;
};
template <typename Bar>
class Foobar
{
public:
using FB = Foo<Bar>;
Foobar(FB::FooPtr fooPtr)
: _fooPtr(fooPtr)
{
}
FB::FooPtr getFooPtr() { return _fooPtr; }
private:
FB::FooPtr _fooPtr;
};
int main()
{
Foobar myFoobar<int>(f::getFooPtr(4));
return myFoobar.getFooPtr()->getBar();
}
When I compile this with g++, I get the following:
dev@ubuntu:~/test/fsm$ g++ -std=c++1z test.cpp
test.cpp:23:23: error: expected ‘)’ before ‘fooPtr’
Foobar(FB::FooPtr fooPtr)
^
test.cpp:28:5: error: need ‘typename’ before ‘Foobar<Bar>::FB::FooPtr’ because ‘Foobar<Bar>::FB’ is a dependent scope
FB::FooPtr getFooPtr() { return _fooPtr; }
^
test.cpp:31:5: error: need ‘typename’ before ‘Foobar<Bar>::FB::FooPtr’ because ‘Foobar<Bar>::FB’ is a dependent scope
FB::FooPtr _fooPtr;
^
test.cpp: In function ‘int main()’:
test.cpp:36:11: error: missing template arguments before ‘myFoobar’
Foobar myFoobar<int>(f::getFooPtr(4));
^
test.cpp:37:11: error: ‘myFoobar’ was not declared in this scope
return myFoobar.getFooPtr()->getBar();
Is the compiler not recognizing the alias FB in the Foobar constuctor? Or is it a problem with resolving Foo::FooPtr to a type?
Thanks!
templates alias
templates alias
asked Nov 13 '18 at 3:13
Moe42
333
asked Nov 13 '18 at 3:13
Moe42
333
asked Nov 13 '18 at 3:13
Moe42
333
asked Nov 13 '18 at 3:13
Moe42
333
asked Nov 13 '18 at 3:13
Moe42
333
333
The error message is clear: "error: need ‘typename’ before ‘Foobar<Bar>::FB::FooPtr’". You have to addtypenamebeforeFB::FooPtr
– max66
Nov 13 '18 at 3:18
You are completely right. I missed it because I was looking at the first error message (from line 23) which does not specify that 'typename' is missing. Adding 'typename' does fix all the cases. Thanks!
– Moe42
Nov 14 '18 at 2:05
add a comment |
The error message is clear: "error: need ‘typename’ before ‘Foobar<Bar>::FB::FooPtr’". You have to addtypenamebeforeFB::FooPtr
– max66
Nov 13 '18 at 3:18
You are completely right. I missed it because I was looking at the first error message (from line 23) which does not specify that 'typename' is missing. Adding 'typename' does fix all the cases. Thanks!
– Moe42
Nov 14 '18 at 2:05
The error message is clear: "error: need ‘typename’ before ‘Foobar<Bar>::FB::FooPtr’". You have to add
typename before FB::FooPtr– max66
Nov 13 '18 at 3:18
The error message is clear: "error: need ‘typename’ before ‘Foobar<Bar>::FB::FooPtr’". You have to add
typename before FB::FooPtr– max66
Nov 13 '18 at 3:18
You are completely right. I missed it because I was looking at the first error message (from line 23) which does not specify that 'typename' is missing. Adding 'typename' does fix all the cases. Thanks!
– Moe42
Nov 14 '18 at 2:05
You are completely right. I missed it because I was looking at the first error message (from line 23) which does not specify that 'typename' is missing. Adding 'typename' does fix all the cases. Thanks!
– Moe42
Nov 14 '18 at 2:05
add a comment |
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The error message is clear: "error: need ‘typename’ before ‘Foobar<Bar>::FB::FooPtr’". You have to add
typenamebeforeFB::FooPtr– max66
Nov 13 '18 at 3:18
You are completely right. I missed it because I was looking at the first error message (from line 23) which does not specify that 'typename' is missing. Adding 'typename' does fix all the cases. Thanks!
– Moe42
Nov 14 '18 at 2:05