c++ template template syntax: simplicity vs useability why not 'auto'

Question

first code below compiled fine after sweating with the word 'class' five times in one line, and definition in "main" of shelf<std::vector, int> top_shelf; looks too fragmented to me, all of that just extract "class E : value_type" out of container.I needed this type_value so I can keep dummy:garbage_value in case an error out of range indexing through the container something like:

  E& return_value_type_at(int x ){ return check(x)? (*container)[x] : garbage_value; }

here is the first code:

#include <iostream>
#include <vector>

template< template<class, class> class C, class E, class A = std::allocator<E> >
class shelf
{
  C<E,A>* container;
// E garbage_value; // not completely implemented
  public:
// bool check(x);
  shelf(C<E,A>& x) : container{&x}{ }
  E& return_value_type_at(int x ){ return /*check(x)?*/(*container)[x]/* : garbage_value*/; }
};

int main()
{
  std::vector<int> box = {1,2,3,4};
  shelf<std::vector,int> top_shelf{box};
  return 0;
}

the second code below, compiled fine, looks a lot simpler:

#include <iostream>
#include <vector>

template<class T>
class shelf
{
  T* container;
  public:
  shelf(T& x) : container{&x}{ }
  auto& value_type_at(int x ){ return (*container)[x]; }
};

int main()
{
  std::vector<int> box = {1,2,3,4};
  shelf< std::vector<int> > top_shelf{box};
  return 0;
}

here the keyword 'auto' helped me out because I have no clue what to replace it with, which is a problem in the same way, how am I going to do the "garbage_value"? another thing why not 'auto' here:

/home/insights/insights.cpp:16:9: error: 'auto' not allowed in template argument
  shelf<auto> top_shelf{box};
        ^~~~

it makes a lot of sense: auto => 'box' template structure.

so is there a way to get "class E" out of the second code?

Ranoiaetep · Accepted Answer · 2022-11-21 05:28:29Z

1

If T is the vector type, then you can get the value type by T::value_type:

template<typename T>
class shelf
{
    T::value_type garbage_value;

    // ⋮
};

answered Nov 21, 2022 at 5:28

Ranoiaetep

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3 Comments

Ammar Tamimi Over a year ago

/home/insights/insights.cpp:8:3: error: missing 'typename' prior to dependent type name 'T::value_type' T::value_type garbage_value; ^~~~~~~~~~~~~

Ranoiaetep Over a year ago

@AmmarTamimi As the error message suggested, you need to have typename prior to T::value_type. Depends on the C++ version you are using, this syntax is required to hint T::value_type here is a type.

Ammar Tamimi Over a year ago

yes thanks, I fixed it last night, now the second code works not only with std::vector, but with std::string.

Richard · Accepted Answer · 2022-11-21 03:20:56Z

1

auto is not allowed to be passed as a template argument.

Instead of using shelf<auto> you can instead use shelf<decltype(box)> as shown below:

shelf<decltype(box)> top_shelf{box}; //works now

edited Nov 21, 2022 at 3:20

answered Nov 21, 2022 at 3:07

Richard

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3 Comments

Ammar Tamimi Over a year ago

I thought: auto top_shelf{box}; will make "top_shelf" a vector<int> type not a "shelf" type.

Richard Over a year ago

@AmmarTamimi You can use shelf<decltype(box)> top_shelf{box};, then top_shelf will be "shelf" as you want. See my updated answer.

Ammar Tamimi Over a year ago

thanks it works, still it has some sort of redundancy.

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