How can I get sizeof a vector::value_type?

Question

I want to get sizeof of the type that is contained in a vector. Here is what I tried:

#include <iostream> #include <vector>  int main() {     std::vector<uint> vecs;     std::cout << sizeof(vecs.value_type) << std::endl;     return 0; } 

From my understanding this should be correct. However, when compiling with GCC 4.8.1 this is what I get:

test-sizeof.cpp: In function ‘int main()’: test-sizeof.cpp:7:27: error: invalid use of ‘std::vector<unsigned int>::value_type’   std::cout << sizeof(vecs.value_type) << std::endl;                            ^

What am I doing wrong? How can I get the size of the contained type?

Answer 1

3.4.3 Qualified name lookup [basic.lookup.qual]

1 The name of a class or namespace member or enumerator can be referred to after the :: scope resolution operator (5.1) applied to a nested-name-specifier that denotes its class, namespace, or enumeration. If a :: scope resolution operator in a nested-name-specifier is not preceded by a decltype-specifier, lookup of the name preceding that :: considers only namespaces, types, and templates whose specializations are types. If the name found does not designate a namespace or a class, enumeration, or dependent type, the program is ill-formed.

In this case, you are accessing a type member from the class template specialization std::vector<uint>, and you need to do it by writing:

std::vector<uint>::value_type 

In case you are actually inside templated code and want to e.g. access the same nested type, you need to prefix it with the keyword typename like this:

typename std::vector<T>::value_type 

In C++11, you can use sizeof(decltype(vecs)::value_type) or also sizeof(decltype(vecs.back())), the latter is convenient if you don't know the precise name of the type but know how to access them through a member function like back().

Note: as pointed out by @Casey in the comments, decltype requires stripping references in order to get the type itself, but for sizeof purposes that doesn't matter.

Answer 2

The member access operator . can only be used to access data members and member functions of classes, not other nested names such as type names. You'll need the scope resolution operator :: to access them, and that can only be applied to the class name (or an alias), not an object of class type:

std::vector<uint>::value_type 

In C++11 or later, decltype can give you a type name, if you have an object and no convenient access to the type:

decltype(vecs)::value_type