c++ - std::shuffle 不能用 std::list 编译

Question

我正在尝试打乱一些生成元素的列表。这是代码:

std::default_random_engine generator (10);
std::list<int> list(10);

int n = 0;
std::generate(list.begin(), list.end(), [&]{ return n++; });
std::shuffle(list.begin(), list.end(), generator);

它不编译。以下是错误:

/include/c++/v1/algorithm:3059:34: Invalid operands to binary expression ('std::__1::__list_iterator<int, void *>' and 'std::__1::__list_iterator<int, void *>')
main.cpp:1:10: In file included from main.cpp:1:

/include/c++/v1/random:1641:10: In file included from /bin/../include/c++/v1/random:1641:

main.cpp:37:10: In instantiation of function template specialization 'std::__1::shuffle<std::__1::__list_iterator<int, void *>, std::__1::linear_congruential_engine<unsigned int, 48271, 0, 2147483647> &>' requested here
/include/c++/v1/iterator:622:1: Candidate template ignored: could not match 'reverse_iterator' against '__list_iterator'
/include/c++/v1/iterator:1017:1: Candidate template ignored: could not match 'move_iterator' against '__list_iterator'
/include/c++/v1/iterator:1369:1: Candidate template ignored: could not match '__wrap_iter' against '__list_iterator'
/include/c++/v1/string:486:11: Candidate template ignored: could not match 'fpos' against '__list_iterator'

有人知道吗？

Answer 1

std::list 不提供对其元素的随机访问，而 std::shuffle() 需要。这就是 std::shuffle() 的签名在其规范中的样子(C++ 标准的第 25.3.12 段):

template<class RandomAccessIterator, class UniformRandomNumberGenerator>
void shuffle(RandomAccessIterator first,
             RandomAccessIterator last,
             UniformRandomNumberGenerator&& g);

如果可以，请考虑改用 std::vector - 顺便说一句，C++ 标准本身鼓励您将其用作默认顺序容器。

例如(live demo on Coliru):

int main()
{
    std::default_random_engine generator(10);
    std::vector<int> v(10);

    std::iota(begin(v), end(v), 0);
    std::shuffle(begin(v), end(v), generator);

    for (auto x : v) { std::cout << x; }
}

std::iota() 算法只是您对 std::generate 的特定用法的一种更简单的替代方法。