c++ - 如何在C++中通过函数序列创建输入迭代器？

Question

我有一个函数，可以获取一对输入迭代器:

template<typename I>
void foo(I begin, I end) {...}

我想用函数生成的序列(例如随机数序列)来填充它。
是否有任何标准的库机制来实现此目的，而无需将序列存储在集合中，然后从集合中获取迭代器？

Answer 1

您将需要自己构建。这是一个如何做的例子:

template<typename F>
class function_iterator {
    using function = F;
public:
    using value_type = std::invoke_result_t<function>;
    using difference_type = std::ptrdiff_t;
    using reference = value_type const&;
    using pointer = value_type const*;
    using iterator_category = std::input_iterator_tag;

    function_iterator(function f, difference_type pos = 0);

    auto operator*() const -> value_type;
    auto operator++() -> function_iterator&;
    auto operator++(int) -> function_iterator;

private:
    function f_;
    difference_type pos_;

    template<typename T_>
    friend bool operator==(function_iterator<T_> const& lhs, function_iterator<T_> const& rhs);
    template<typename T_>
    friend bool operator!=(function_iterator<T_> const& lhs, function_iterator<T_> const& rhs);
};

template<typename T>
function_iterator<T>::function_iterator(function f, difference_type pos) :
f_(f), pos_(pos) {
}

template<typename F>
auto function_iterator<F>::operator*() const -> value_type {
    return f_();
}
template<typename F>
auto function_iterator<F>::operator++() -> function_iterator& {
    ++pos_;
    return *this;
}
template<typename F>
auto function_iterator<F>::operator++(int) -> function_iterator {
    auto it = *this;
    ++*this;
    return it;
}

template<typename T_>
bool operator==(function_iterator<T_> const& lhs, function_iterator<T_> const& rhs) {
    return lhs.pos_ == rhs.pos_;
}
template<typename T_>
bool operator!=(function_iterator<T_> const& lhs, function_iterator<T_> const& rhs) {
    return !(lhs == rhs);
}

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