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std::reverse_iterator<Iter>:: operator*,->

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C++
Iterator library
Iterator concepts
Iterator primitives
Algorithm concepts and utilities
Indirect callable concepts
Common algorithm requirements
(C++20)
(C++20)
(C++20)
Utilities
(C++20)
Iterator adaptors
Range access
(C++11) (C++14)
(C++14) (C++14)
(C++11) (C++14)
(C++14) (C++14)
(C++17) (C++20)
(C++17)
(C++17)
std::reverse_iterator
reference operator * ( ) const ;
(1) (自 C++17 起为 constexpr)
(2)
pointer operator - > ( ) const ;
(自 C++17 起为 constexpr)
(直至 C++20)
constexpr pointer operator - > ( ) const

requires ( std:: is_pointer_v < Iter > ||

requires ( const Iter i ) { i. operator - > ( ) ; } ) ;
(自 C++20 起)

返回指向 current 前一个元素的引用或指针。

成员函数 等效于
operator * Iter tmp = current ; return *-- tmp ;
operator - >

return & ( operator * ( ) ) ;

(C++11 前)

return std:: addressof ( operator * ( ) ) ;

(C++11 起)
(C++20 前)
(C++20 起)

目录

返回值

如上所述。

示例

运行此代码 
#include <complex>
#include <iostream>
#include <iterator>
#include <vector>
int main()
{
    using RI0 = std::reverse_iterator<int*>;
    int a[]{0, 1, 2, 3};
    RI0 r0{std::rbegin(a)};
    std::cout << "*r0 = " << *r0 << '\n';
    *r0 = 42;
    std::cout << "a[3] = " << a[3] << '\n';
    using RI1 = std::reverse_iterator<std::vector<int>::iterator>;
    std::vector<int> vi{0, 1, 2, 3};
    RI1 r1{vi.rend() - 2};
    std::cout << "*r1 = " << *r1 << '\n';
    using RI2 = std::reverse_iterator<std::vector<std::complex<double>>::iterator>;
    std::vector<std::complex<double>> vc{{1, 2}, {3, 4}, {5, 6}, {7, 8}};
    RI2 r2{vc.rbegin() + 1};
    std::cout << "vc[2] = (" << r2->real() << ',' << r2->imag() << ")\n";
}

输出： 

*r0 = 3
a[3] = 42
*r1 = 1
vc[2] = (5,6)

缺陷报告

下列行为变更缺陷报告被追溯应用于先前发布的 C++ 标准。

缺陷报告 适用范围 发布时行为 正确行为
LWG 2188 C++11 operator - > 使用 & 取地址 改用 std::addressof

参见

通过索引访问元素
(公开成员函数)