对于具有默认分配器的标准容器，std :: container :: size_type是否保证为size_t? [英] Is std::container::size_type guaranteed to be size_t for standard containers with default allocator?

问题描述

赞:

• std::string<T>::size_type
• std::list<T>::size_type
• std::map<T>::size_type
• std::vector<T>::size_type
• 等

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

cplusplus.com 和

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?

推荐答案

对于STL容器-不. [container.requirements.general]中标准的表96列出了任何容器X的容器要求，很清楚地说明了这一点:

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:

但是，对于basic_string，size_type被定义为

typedef typename allocator_traits<Allocator>::size_type size_type;

，对于std::allocator<..>作为分配器，它将依次为size_t.

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

此外，根据[array.overview]/3，std::array使用size_t作为size_type.

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

这篇关于对于具有默认分配器的标准容器，std :: container :: size_type是否保证为size_t?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！