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auto specifier (C++11 起)

来自cppreference.com

 
 
C + +语言
大会的主题
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General topics
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流量控制
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Flow control
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条件执行语句
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Conditional execution statements
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迭代语句
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Iteration statements
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跳转语句
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Jump statements
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功能
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Functions
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函数声明
lambda函数的声明
函数模板
的历史。内嵌说明
异常规范 (过时了)
noexcept说明 (C++11)
例外
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Exceptions
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命名空间
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Namespaces
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类型
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Types
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decltype specifier (C++11)
规范
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Specifiers
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CV符
存储时间说明符
constexpr说明 (C++11)
汽车符 (C++11)
alignas说明 (C++11)
初始化
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Initialization
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Literals
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布尔文字
nullptr (C++11)
用户定义的 (C++11)
表达式
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Expressions
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另一种表示形式
实用工具
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Utilities
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类型
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Types
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typedef declaration
声明类型别名 (C++11)
属性 (C++11)
施放
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Casts
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隐式转换
const_cast conversion
static_cast conversion
dynamic_cast conversion
reinterpret_cast conversion
C-风格和功能转换
内存分配
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Memory allocation
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Classes
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类特定的功能特性
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Class-specific function properties
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虚函数
覆盖说明 (C++11)
最后说明 (C++11)
明确的 (C++11)
静态的
特殊的成员函数
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Special member functions
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模板
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Templates
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类模板
函数模板
模板特化
参数包 (C++11)
杂项
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Miscellaneous
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内联汇编
 
指定被声明的变量的类型将被自动推导出它的初始化。对于函数,指定的返回类型是一个后返回类型.
Original:
Specifies that the type of the variable that is being declared will be automatically deduced from its initializer. For functions, specifies that the return type is a trailing return type.
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目录

[编辑] 语法

auto variable initializer (C++11 起)
auto function -> return type (C++11 起)

[编辑] 解释

1)
声明变量时,在块的范围,在命名空间范围内,在初始化语句的for循环等,被忽略的变量的类型和关键字auto可以用来代替.
Original:
When declaring variables in block scope, in namespace scope, in init statements of for loops, etc, the type of the variable may be omitted and the keyword auto may be used instead.
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的类型初始值设定项已被确定后,编译器决定,将取代关键字auto,如果使用的是从一个函数调用模板参数推导规则的类型。关键字auto可能会伴随着修改,如const&,这将参与的类型推演。例如,给定const auto& i = expr;,类型i是完全相同的参数的类型,在一个假想的模板utemplate<class U> void f(const U& u)如果函数调用f(expr)编译.
Original:
Once the type of the initializer has been determined, the compiler determines the type that will replace the keyword auto as if using the rules for template argument deduction from a function call. The keyword auto may be accompanied by modifies, such as const or &, which will participate in the type deduction. For example, given const auto& i = expr;, the type of i is exactly the type of the argument u in an imaginary template template<class U> void f(const U& u) if the function call f(expr) was compiled.
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2)
函数声明,关键字auto不执行自动检测的类型。它只是作为后返回类型语法的一部分.
Original:
In a 函数声明, the keyword auto does not perform automatic type detection. It only serves as a part of the trailing return type syntax.
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[编辑] 注释

直到C + +11,auto的语义的存储时间说明.
Original:
Until C++11, auto had the semantic of a 存储时间说明.
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[编辑] 为例

#include <iostream>
#include <cmath>
#include <typeinfo>
 
template<class T, class U>
auto add(T t, U u) -> decltype(t + u) // the return type of add is the type of operator+(T,U)
{
    return t + u;
}
 
auto get_fun(int arg)->double(*)(double) // same as double (*get_fun(int))(double)
{
    switch (arg) {
        case 1: return std::fabs;
        case 2: return std::sin;
        default: return std::cos;
    }
}
 
int main()
{
    auto a = 1 + 2;
    std::cout << "type of a: " << typeid(a).name() << '\n';
    auto b = add(1, 1.2);
    std::cout << "type of b: " << typeid(b).name() << '\n';
    //auto int c; //compile-time error
    auto d = {1, 2};
    std::cout << "type of d: " << typeid(d).name() << '\n';
 
    auto my_lambda = [](int x) { return x + 3; };
    std::cout << "my_lambda: " << my_lambda(5) << '\n';
 
    auto my_fun = get_fun(2);
    std::cout << "type of my_fun: " << typeid(my_fun).name() << '\n';
    std::cout << "my_fun: " << my_fun(3) << '\n';
}

Output:

type of a: int
type of b: double
type of d: std::initializer_list<int>
my_lambda: 8
type of my_fun: double (*)(double)
my_fun: 0.14112