std :: size_t或std :: vector< Foo> :: size_type? [英] std::size_t or std::vector<Foo>::size_type?

问题描述

当我在std::vector<Foo>(或每个具有随机访问迭代器的容器)上循环时，我使用无符号整数变量i.如果要遵守规范，应该使用std::size_t还是容器本身指定的类型:std::vector<Foo>::size_type吗?

When I loop on a std::vector<Foo> (or every container having random access iterator) I use an unsigned integer variable i. If I want to respect the norm, should I use std::size_t or the type given by the container itself : std::vector<Foo>::size_type ?

如果出于可读性原因选择了std::size_t，可以确定std名称空间中每个容器的每个实现都将std::size_t用作size_type吗?

If I chose std::size_t (for readability reasons), can I be sure that every implementation of every container in std namespace uses std::size_t as size_type ?

注意:出于兼容性原因，我仅使用C ++ 98.

Note : I use C++98 only (for compatibility reasons).

推荐答案

std::vector<Foo>::size_type与std::size_t相同并不是不必要.即使对于C ++ 11也是如此.

It is not necessarily true that std::vector<Foo>::size_type is the same as std::size_t. This is true even for C++11.

但就我个人而言，无论类型如何，我都将std::size_t用于std::vector索引.

But personally I use std::size_t for a std::vector index irrespective of the type.

如果您特别勤奋，则可以始终使用静态断言.显然，static_assert是C ++ 98以后的新增内容，但是按照该标准，您可以使用类似

You could always use a static assertion if you're feeling particularly diligent. Obviously static_assert is a later addition beyond what's in C++98, but in that standard you could use something like

static char wrong_size_1[1 + sizeof(std::size_t) - sizeof(std::vector<Foo>::size_type)];

static char wrong_size_2[1 - sizeof(std::size_t) + sizeof(std::vector<Foo>::size_type)];

如果类型类型的大小不同，则会导致编译时失败.

which would induce compile time failures if type types are not the same size.

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