Is std::container::size_type guaranteed to be size_t for standard containers with default allocator?

问题

Like:

• std::string<T>::size_type
• std::list<T>::size_type
• std::map<T>::size_type
• std::vector<T>::size_type
• etc.

Both cplusplus.com and cppreference.com say that they are usually size_t, but are they truly, unambiguously guaranteed by the standard to be size_t unless a custom allocator is used?

回答1:

For STL-containers - nope. Table 96 of the standard in [container.requirements.general], which lists container requirements for any container X, explains it pretty clear:

---

However, for basic_string, size_type is defined as

typedef typename allocator_traits<Allocator>::size_type size_type;

which in turn will be size_t for std::allocator<..> as the allocator.

Also, std::array uses size_t as size_type, according to [array.overview]/3.

回答2:

size_type isn't guaranteed to be size_t.

But the default allocator size_type is, so the default is size_t.

From the standard 20.6.9

template <class T> class allocator {
public:
typedef size_t size_type;
typedef ptrdiff_t difference_type;
....

The container's size_type is derived from the allocator:

typedef typename allocator_traits<Allocator>::size_type size_type;

来源：https://stackoverflow.com/questions/26433802/is-stdcontainersize-type-guaranteed-to-be-size-t-for-standard-containers-wit