Question

我的问题是安全问题。我搜索了cplusplus.com和cppreference.com，他们似乎缺乏std :: move期间的迭代器安全性。具体来说：使用已移动对象的迭代器调用std :: unordered_map :: erase（iterator）是否安全？示例代码：

#include <unordered_map>
#include <string>
#include <vector>
#include <iostream>
#include <memory>

class A {
public:
    A() : name("default ctored"), value(-1) {}
    A(const std::string& name, int value) : name(name), value(value) { }
    std::string name;
    int value;
};
typedef std::shared_ptr<const A> ConstAPtr;

int main(int argc, char **argv) {
    // containers keyed by shared_ptr are keyed by the raw pointer address
    std::unordered_map<ConstAPtr, int> valued_objects;

    for ( int i = 0; i < 10; ++i ) {
        // creates 5 objects named "name 0", and 5 named "name 1"
        std::string name("name ");
        name += std::to_string(i % 2);

        valued_objects[std::make_shared<A>(std::move(name), i)] = i * 5;
    }

    // Later somewhere else we need to transform the map to be keyed differently
    // while retaining the values for each object

    typedef std::pair<ConstAPtr, int> ObjValue;

    std::unordered_map<std::string, std::vector<ObjValue> > named_objects;

    std::cout << "moving..." << std::endl;

    // No increment since we're using .erase() and don't want to skip objects.
    for ( auto it = valued_objects.begin(); it != valued_objects.end(); ) {
        std::cout << it->first->name << "\t" << it->first.value << "\t" << it->second << std::endl;

        // Get named_vec.
        std::vector<ObjValue>& v = named_objects[it->first->name];
        // move object :: IS THIS SAFE??
        v.push_back(std::move(*it));

        // And then... is this also safe???
        it = valued_objects.erase(it);
    }

    std::cout << "checking... " << named_objects.size() << std::endl;
    for ( auto it = named_objects.begin(); it != named_objects.end(); ++it ) {
        std::cout << it->first << " (" << it->second.size() << ")" << std::endl;
        for ( auto pair : it->second ) {
            std::cout << "\t" << pair.first->name << "\t" << pair.first->value << "\t" << pair.second << std::endl;
        }
    }

    std::cout << "double check... " << valued_objects.size() << std::endl;
    for ( auto it : valued_objects ) {
        std::cout << it.first->name << " (" << it.second << ")" << std::endl;
    }

    return 0;
}

我问的原因是，从unordered_map的迭代器中移动对可能（？）因此*重新移动迭代器存储的键值，因此使其散列无效;因此，之后的任何操作都可能导致未定义的行为。除非那不是这样吗？

我认为值得注意的是，上述内容似乎已成功按照GCC 4.8.2的规定运行，所以我希望看看我是否错过了支持或明确不支持该行为的文档。

Answer 1

// move object :: IS THIS SAFE??
v.push_back(std::move(*it));

是的，它是安全的，因为它实际上并没有修改密钥。它不能，因为键是const。 *it的类型为std::pair<const ConstAPtr, int>。移动时，第一个成员（const ConstAPtr）实际上没有移动。它会被std::move转换为r值，并变为const ConstAPtr&&。但这与移动构造函数不匹配，后者需要非const ConstAPtr&&。所以改为调用复制构造函数。

将对象从一个unordered_map移动到另一个容器

1 个答案: