Can conditional parameters be passed to a template? c++ - c++

I am writing a template to abstract a buffer class. Can I pass conditional_variables as parameters to the template or do they need to be global? What is the best way to template this to abstract it without explicitly 'hardcoding' it in?

You can use references to them:
template<std::condition_variable &> class F;
(or pointers, analogously). But this actually makes very little sense for the arguments to such a template should be very static and compile-time known, i.e. practically the very same globals.

Related

Template class forward declaration with default argument [duplicate]

Suppose we have a class template with default template parameter:
template <typename T = int>
class Foo {};
We can omit angle brackets when creating a variable inside a function:
int main()
{
Foo a; // gets properly deduced as Foo<int>
}
But we can't do that for member variables:
struct S
{
Foo a; // Deduce Foo<int>
};
We can't have derivative types such as this:
Foo* ptr; // Foo<int>*
Foo& ref; // Foo<int>&
int Foo::* mem_ptr; // int Foo<int>::*
std::function<Foo(const Foo&)> fn; // std::function<Foo<int>(const Foo<int>&)>
We can't accept parameters and return them:
Foo Bar(const Foo&); // Foo<int> (*)(const Foo<int>&)
Why? Is this considered a bug in the standard? Is there a proposal to fix it? Are there any actual problems with omitting angle brackets?
My use case:
I have a class template which provides default argument. The template parameter is an expert-only feature that I myself never use but it is there for those 1% of experts who want that total flexibility. Now for other 99% I want to hide the fact that Foo is actually a class template but it doesn't work because users have to type Foo<> when declaring it as a member variable, current solution is this:
template <typename T = int>
class BasicFoo {};
using Foo = BasicFoo<>;
But it complicates implementation code and is not elegant at all.
Is this considered a bug in the standard?
No.
Templates are a named construct which generates another construct (classes/functions/variables) based on a set of parameters. The name of a template is not the name of the construct which it generates. The name of a template is just the name of the template; to name the thing the template generates, you must provide the template parameters.
Foo is the name of a template; Foo<> is the name of a class generated by that template and its associated template parameters.
There are a couple of places where C++ allows a template to be used in such a way that its parameters are deduced from a sequence of expressions. But these are very specific places, created for convenience purposes. They do not exist for the purpose of hiding the fact that a name represents a template rather than the generated construct.
Is there a proposal to fix it?
There is nothing broken to fix. And there are at present no proposals adding changes in this way.
Are there any actual problems with omitting angle brackets?
Define "actual problem". Is it theoretically possible to have the language altered so that, if all of a template's parameters are defaulted, the name of the template can be used without template parameters to simultaneously mean the template and the thing the template generates?
It is probably possible. But it would be complicated to specify. You would need a serious spec-doctor, one who understands the C++ grammar at a deep level, to know for sure whether it is possible, and what exactly would need to be changed to do it.
But at the end of the day, it would only ever be useful for a small, select set of templates: templates that have default values for all of its parameters. The real question is whether that is a common enough case to be worth the effort.
I don't consider myself a language expert, but my stance would be that the problem your proposal tries to tackle is solved much simpler in the same way std::(basic_)string does it.
We have
template<
class CharT,
class Traits = std::char_traits<CharT>,
class Allocator = std::allocator<CharT>
> class basic_string;
and then a set of typedefs for "non-expert" users, such as std::string for std::basic_string<char>.
If an expert user wants to make use of the other template parameters, they can themselves define a type alias, which is nice and coherent with the above. Moreover, this cleanly separates the template from the types that are created from it.
Your suggestion of allowing templates with defaults for all parameters to be named by MyTemplate alone, instead of requiring MyTemplate<>, or making use of using MyTemplate = MyBasicTemplate<>;, has the following issues:
It complicates the grammar and specification of the language. You need to touch the allowed syntax of all the contexts mentioned in your question, adding the ability to use a template name where a type name would be expected, but only if the relevant template has default values for all template parameters. And if you don't change all of them, you introduce weirdly inconsistent behavior.
There is some overlap between your suggestion and CTAD, but CTAD is decidedly about reducing type verbosity for initialization. CTAD offers significant comfort within its scope and is extensible through deduction guides, whereas your proposal's syntactic sugar is only relevant in a tiny usage niche, with much smaller benefits.
There is the danger of accidentally using the wrong template parameters (did you mean the default template parameters or did you just forget to specify the ones you wanted?). Even if that is not a concern in your use case, the standard would have to concern itself with that potential issue.
There is also the danger of your suggestion conflicting with deduction guides. Who should win?
Your problem is easily and conveniently solved with existing language tools (see above). I disagree that this "complicates" implementation code (complexity is literally only increased by a single typedef/using, (re)naming your template is absolutely trivial work) or that it is inelegant.
Overall, the problem you intend to solve (saving library implementers a using, or users a <> (or using), exclusively for all-defaulted templates) is fringe at best and will not be a sufficient motivation for significantly altering several core aspects the language. That's my prediction at least.

Binding times for template variables

In C++ for a template T variable for a class.
What would it's binding time be?
I'm thinking compile time because it would need to know what value is being passed to it before binding it to a value/address.
Yes. You could think of templates as "class factories", the result of using a template is a class that has been specialized for the specific types you use in the template. So this is all compile-time specialization.

(Can I minimize the) Footprint of template in my C++ Code?

I have a large class which basically handles one buffer of variable (numeric) datatype. So it seems a good choice to use a class template with this datatype as the only parameter. I'm not experienced in C++ and I wonder/worry a bit about the "footprint" such a template makes in my code.
There are three implications of templates which in my (C++ unexperienced) eyes are not necessary and make code ugly. I tried to avoid them, but neither did I find a good example how to do it nor did I manage to find it out by myself.
So the goal of this question is: Can you either confirm the following statements or give a counterexample?
When using a class template, all class methods have to go into the header file. Even if they have no templated type in their interface or implementation.
When using a static method or member of the class, I always have to specify a template parameter (MyClass< double > :: MY_STATIC), even if the templatization does not affect any of the static properties of the class.
When using the class as a parameter for a function, I always have to give a template parameter, even when this function does not access any of the templated members? (function myFunc(MyClass< double> & myClass){ do something } )
As a general rule, don't have functions/data members in a template class which does not use the template parameters. Have a base class, put all non-template related things there, your template class should derive from it.
To answer your questions:
yes, everywhere where you need to instantiate the template, you need to see the full definition of the class and it's functions
yep, but put that into the base class
yes, see above
EDIT: One of the reasons to move to base class is code bloating (this expression actually exist, you can google it for more info): If you don't move the template unrelated code to a base class, the very same template independent code will be copied for all instantiation of your template, which means a lot of unnecessary code. If you put it to a base class, you will only have this code once.
Yes. On the plus side, the code is only generated when the metod is actually used for the specialization.
Yes. However, there is no (other then design choice) need for a static method to be a memeber of the templated class if it has no use for the templated parameter.
Yes. The size and memory layout of the structure is determined by the template parameter.

Must an unused function in a template class be valid?

In a template I have some functions which are only valid for certain template types. GCC seems to be happy with this, but I'm not sure it is valid. Unlike typical SFINAE the function itself is not a template.
template<typename T>
struct generic {
T item;
void get_limited() {
item.limited();
}
};
Provided I don't call get_limited, is it okay to instantiate this class with a type that does not implement limited?
If no, how can I solve this? I have a generic container class where certain features are enabled based on the allocate it is storing (so not directly on the type as above, but still a template parameter).
Template functions are instantiated on demand, so if there is no use of the function it need not be correct, at least for some possible instantiations. The standard does state that if a template is not valid for any instantiating type, the program is ill-formed (although the compiler is not required to diagnose it) even if it is never instantiated.
This feature is used in the standard library in different parts, where the requirements of a single function might be stricter than the general requirements that the template places on the instantiating types. For example, in the case of std::map, in general, the value type does not need to be default constructible, but if you want to use operator[] then it needs to be, since that operator might need to create an empty element if the key is not present.

C++ std::vector with pointers to a template class

In a project I am working on I am tying to make a vector with pointers to a template class.
template <typename T>
std::vector<templateClass<T>*> vec;
However, this gives me two errors:
Error C2133: vec : unknown size
Error C2998: std::vector<templateClass<T>*> vec : cannot be a template definition
If I change the code to:
std::vector<templateClass<int>*> vec;
It works fine, so I guess the problem isn't that you cant use template classes with vectors, but that you need to tell the compiler what type to use. Is there any way around this?
When you create a class instance you have to choose the type. In the definition you can write T but at the moment of create instance you have to specify the type.
So if you want define and not creating an instance use typedef.
You can't have a templated member.
The template must come from the class or function templated declaration.
template <typename T>
class blah {
std::vector<templateClass<T>*> vec;
}
The compiler needs the templated typename to be defined somewhere in the code, for example: blah<int>
If you'd have a templated member, you couldn't define the type anywhere in the code, and the compiler wouldn't be able to decide the member's type.
The templated typename is decided when you first use the function or class (either explicitly or implicitly) so you'd have to have the template definition and implementation somewhere that is accessible by the calling code.
It looks like you are trying to define a new type vec<T> as a shortcut to the longer templetized expression. Normally, this would be done with a typedef, however C++ does not support templetized typedefs.
Note, that in the current code you are essentially trying to define a variable called vec, but you are not giving it a specific type for T and that's why the compiler is complaining.
Currently C++ doesn’t support template typedefs so you have to use the most common solution, proposed by Herb Sutter (http://gotw.ca/gotw/079.htm)