This question already has answers here:
templates: parent class member variables not visible in inherited class
(3 answers)
Closed 7 years ago.
The following code cannot compile - use of undeclared identifier. I use GCC and XCode for compilation.
Everything is in a single header file.
include "MyArray.h"
template <typename T>
class MyBase {
public:
MyBase();
virtual ~MyBase();
void addStuff(T* someStuff);
protected:
MyArray<T*> stuff;
};
template <typename T>
MyBase<T>::MyBase() {}
template <typename T>
MyBase<T>::~MyBase() {}
template <typename T>
void MyBase<T>::addStuff(T* someStuff) {
stuff.add(someStuff);
}
// ---------------------
template <typename T>
class MyDerived : public MyBase<T> {
public:
MyDerived();
virtual ~MyDerived();
virtual void doSomething();
};
template <typename T>
MyDerived<T>::MyDerived() {}
template <typename T>
MyDerived<T>::~MyDerived() {}
template <typename T>
void MyDerived<T>::doSomething() {
T* thingy = new T();
addStuff(thingy); //here's the compile error. addStuff is not declared.
}
Does anyone have an explanation?
Thanks in advance!
try
this->addStuff(thingy);
It's due to template inheritance. In such case you should mannualy specify using for base methods:
template <typename T>
MyDerived<T>::doSomething() {
using MyBase<T>::addStuff;
T* thingy = new T();
addStuff(thingy);
}
or do it by this pointer:
template <typename T>
MyDerived<T>::doSomething() {
T* thingy = new T();
this->addStuff(thingy);
}
There are several issues:
Missing semicolons after class definitions.
Missing type for doSomething method declaration/definition.
Missing type for definition of addStuff method.
After fixing that it seems to work.
Edit: As you have fixed the syntax errors and it still does not work. As others have suggested your compiler may require you to call the addStuff method with this-> prefix:
this->addStuff(thingy);
use this pointer to invoke the addStuff method i.e
this->addStuff(thingy);
Related
I can't seem to call a method of a base class without scoping to the base class, and it seems that this is because I have overloaded the method. If I do not overload the method then the compiler doesn't complain. Here's an example of what I'd like to do:
struct BaseClass {
template <typename T> T& foo(T& t) {
return t;
}
};
class ChildClass: public BaseClass {
public:
// The presence of this template causes compiler confusion
template <class T> T& foo(T& t, int szl) {
return t;
}
template <class T> int bar(T& t) {
// But if I scope this as BaseClass::foo(...) it's all good
return foo(t);
}
};
int main() {
int t = 1;
ChildClass c;
c.bar(t);
}
If in bar(...) I call BaseClass::foo(...) the compiler does not complain, but I don't see any ambiguity here and so I'm confused as to why I'd need to do this.
When the compiler tries to match a function name with a function, it does so in two steps. In the first step it finds all the functions that match the given name. If it finds more than one function, it tries the logic of overload resolution to find the best matching function.
In the first step, if the compiler finds a name in the class, it stops looking for functions of the same name in base classes. In your case, since it finds a foo in ChildClass, it stops searching for functions named foo in BaseClass. However, the only matching foo does not match the call and the compiler reports an error.
How to resolve the problem:
Use the method you described in your post. Call BaseClass::foo(...).
Bring all the foo from BaseClass into the scope of ChildClass.
class ChildClass: public BaseClass {
public:
using BaseClass::foo;
template <class T> T& foo(int baz, T& t, int szl) {
return t;
}
template <class T> int bar(T& t) {
return sizeof(foo(1, t)); // Should work.
}
};
From the C++ standard 3.3.10 paragraph 1 :
A name can be hidden by an explicit declaration of that same name in
a nested declarative region or derived class
ChildClass::foo shadows the definition of BaseClass::foo. All you need to bring it into scope is a using directive:
class ChildClass: public BaseClass {
public:
using BaseClass::foo;
template <class T> T& foo(int baz, T& t, int szl);
};
You need a using, like:
using BaseClass::foo;
I can't seem to call a method of a base class without scoping to the base class, and it seems that this is because I have overloaded the method. If I do not overload the method then the compiler doesn't complain. Here's an example of what I'd like to do:
struct BaseClass {
template <typename T> T& foo(T& t) {
return t;
}
};
class ChildClass: public BaseClass {
public:
// The presence of this template causes compiler confusion
template <class T> T& foo(T& t, int szl) {
return t;
}
template <class T> int bar(T& t) {
// But if I scope this as BaseClass::foo(...) it's all good
return foo(t);
}
};
int main() {
int t = 1;
ChildClass c;
c.bar(t);
}
If in bar(...) I call BaseClass::foo(...) the compiler does not complain, but I don't see any ambiguity here and so I'm confused as to why I'd need to do this.
When the compiler tries to match a function name with a function, it does so in two steps. In the first step it finds all the functions that match the given name. If it finds more than one function, it tries the logic of overload resolution to find the best matching function.
In the first step, if the compiler finds a name in the class, it stops looking for functions of the same name in base classes. In your case, since it finds a foo in ChildClass, it stops searching for functions named foo in BaseClass. However, the only matching foo does not match the call and the compiler reports an error.
How to resolve the problem:
Use the method you described in your post. Call BaseClass::foo(...).
Bring all the foo from BaseClass into the scope of ChildClass.
class ChildClass: public BaseClass {
public:
using BaseClass::foo;
template <class T> T& foo(int baz, T& t, int szl) {
return t;
}
template <class T> int bar(T& t) {
return sizeof(foo(1, t)); // Should work.
}
};
From the C++ standard 3.3.10 paragraph 1 :
A name can be hidden by an explicit declaration of that same name in
a nested declarative region or derived class
ChildClass::foo shadows the definition of BaseClass::foo. All you need to bring it into scope is a using directive:
class ChildClass: public BaseClass {
public:
using BaseClass::foo;
template <class T> T& foo(int baz, T& t, int szl);
};
You need a using, like:
using BaseClass::foo;
This question already has answers here:
C++ syntax for explicit specialization of a template function in a template class?
(9 answers)
Closed 8 years ago.
template<typename T>
class CConstraint
{
public:
CConstraint()
{
}
virtual ~CConstraint()
{
}
template <typename TL>
void Verify(int position, int constraints[])
{
}
template <>
void Verify<int>(int, int[])
{
}
};
Compiling this under g++ gives the following error:
Explicit specialization in non-namespace scope 'class CConstraint'
In VC, it compiles fine. Can anyone please let me know the workaround?
VC++ is non-compliant in this case - explicit specializations have to be at namespace scope. C++03, §14.7.3/2:
An explicit specialization shall be declared in the namespace of which the template is a member, or, for member templates, in the namespace of which the enclosing class or enclosing class template is a member.
An explicit specialization of a member function, member class or static data member of a class template shall be declared in the namespace of which the class template is a member.
Additionally you have the problem that you can't specialize member functions without explicitly specializing the containing class due to C++03, §14.7.3/3, so one solution would be to let Verify() forward to a, possibly specialized, free function:
namespace detail {
template <typename TL> void Verify (int, int[]) {}
template <> void Verify<int>(int, int[]) {}
}
template<typename T> class CConstraint {
// ...
template <typename TL> void Verify(int position, int constraints[]) {
detail::Verify<TL>(position, constraints);
}
};
Another way to solve it is by delegating to a private function and overloading that function. This way you still have access to member data of *this and to the outer template parameter type.
template<typename T>
struct identity { typedef T type; };
template<typename T>
class CConstraint
{
public:
template <typename TL>
void Verify(int position, int constraints[])
{
Verify(position, constraints, identity<TL>());
}
private:
template<typename TL>
void Verify(int, int[], identity<TL>)
{
}
void Verify(int, int[], identity<int>)
{
}
};
Just take the template specialization outside the class declaration.
gcc doesn't allow inline template specialization.
As another option, just deleting line
template<>
seems to work for me.
Even better: you can combine partial specialisation with default template arguments. This way modification to the VC++ code are minor, because calls to the specialised function do not need to be modified.
template <typename TL, class Dummy=int>
void Verify(int position, int constraints[])
{
}
template <class Dummy=int>
void Verify<int, Dummy>(int, int[])
{
}
You may not be able to explicitly specialize the member template, but you can partially specialize it. If you add a second parameter "int dummyParam" and also add it to the specialization, it should work with both compilers.
Not that I knew this more than 10 seconds ago, but googling on the same error, I ran into this link and it worked for my member template specialization.
First, I hope the question title makes somehow sense.
I have the following class structure:
class A : public Singleton<A>
{
public:
template <typename T> class Buffer
{
public:
//ctor & dtor
T* get() { return ptr; }
private:
T* ptr;
};
// class A stuff
};
This class should be passed to a handler class as template argument and used within the template functions of the class:
template <class MODEL> class Handler
{
public:
// ctor & dtor
template <typename T> typename MODEL::Buffer<T>* create(...) // error c2988
{ // create a buffer }
};
The compiler however is unable to determine the correct typenames and throws error C2988: unrecognizable template declaration/definition (vc++ Nov 2012 CTP compiler within VS 2012). I cannot find any solution to tell the compiler how to handle the innerclass template. So the question is: How to make this work?
Any help would be highly appreciate.
You need to help the compiler to disambiguate and tell it what Buffer is:
template <typename T> typename MODEL::template Buffer<T>* create(...);
// ^^^^^^^^
See this Q&A for more info.
just now I had to dig through the website to find out why template class template member function was giving syntax errors:
template<class C> class F00 {
template<typename T> bar();
};
...
Foo<C> f;
f.bar<T>(); // syntax error here
I now realize that template brackets are treated as relational operators. To do what was intended the following bizarre syntax is needed, cf Templates: template function not playing well with class's template member function:
f.template bar<T>();
what other bizarre aspects and gotcha of C++/C++ templates you have encountered that were not something that you would consider to be common knowledge?
I got tripped up the first time I inherited a templated class from another templated class:
template<typename T>
class Base {
int a;
};
template<typename T>
class Derived : public Base<T> {
void func() {
a++; // error! 'a' has not been declared
}
};
The problem is that the compiler doesn't know if Base<T> is going to be the default template or a specialized one. A specialized version may not have int a as a member, so the compiler doesn't assume that it's available. But you can tell the compiler that it's going to be there with the using directive:
template<typename T>
class Derived : public Base<T> {
using Base<T>::a;
void func() {
a++; // OK!
}
};
Alternatively, you can make it explicit that you are using a member of T:
void func {
T::a++; // OK!
}
This one got me upset back then:
#include <vector>
using std::vector;
struct foo {
template<typename U>
void vector();
};
int main() {
foo f;
f.vector<int>(); // ambiguous!
}
The last line in main is ambiguous, because the compiler not only looks up vector within foo, but also as an unqualified name starting from within main. So it finds both std::vector and foo::vector. To fix this, you have to write
f.foo::vector<int>();
GCC does not care about that, and accepts the above code by doing the intuitive thing (calling the member), other compilers do better and warn like comeau:
"ComeauTest.c", line 13: warning: ambiguous class member reference -- function
template "foo::vector" (declared at line 8) used in preference to
class template "std::vector" (declared at line 163 of
"stl_vector.h")
f.vector<int>(); // ambiguous!
The star of questions about templates here on SO: the missing typename!
template <typename T>
class vector
{
public:
typedef T * iterator;
...
};
template <typename T>
void func()
{
vector<T>::iterator it; // this is not correct!
typename vector<T>::iterator it2; // this is correct.
}
The problem here is that vector<T>::iterator is a dependent name: it depends on a template parameter. As a consequence, the compiler does not know that iterator designates a type; we need to tell him with the typename keyword.
The same goes for template inner classes or template member/static functions: they must be disambiguated using the template keyword, as noted in the OP.
template <typename T>
void func()
{
T::staticTemplateFunc<int>(); // ambiguous
T::template staticTemplateFunc<int>(); // ok
T t;
t.memberTemplateFunc<int>(); // ambiguous
t.template memberTemplateFunc<int>(); // ok
}
Out of scope class member function definition:
template <typename T>
class List { // a namespace scope class template
public:
template <typename T2> // a member function template
List (List<T2> const&); // (constructor)
…
};
template <typename T>
template <typename T2>
List<T>::List (List<T2> const& b) // an out-of-class member function
{ // template definition
…
}