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

来自cppreference.com
 
 
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(C++20)
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(C++20)
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(C++20)
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范围访问
(C++11)(C++14)
(C++14)(C++14)  
(C++11)(C++14)
(C++14)(C++14)  
(C++17)(C++20)
(C++17)
(C++17)
 
 
(1)
reference operator*() const;
(C++17 前)
constexpr reference operator*() const;
(C++17 起)
(2)
pointer operator->() const;
(C++17 前)
constexpr pointer operator->() const;
(C++17 起)
(C++20 前)
constexpr pointer operator->() const

    requires (std::is_pointer_v<Iter> ||

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

返回指向 current 前一元素的引用或指针。

1) 等价于 Iter tmp = current; return *--tmp;
2) 等价于 return std::addressof(operator*());
(C++20 前)
2)Iter 为指针类型则等价于 return current - 1;。否则等价于 return std::prev(current).operator->();
(C++20 起)

参数

(无)

返回值

指向 current 前一元素的引用或指针。

示例

#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)

参阅

按索引访问元素
(公开成员函数)