如何将智能指针向量转换为包含指向常量的智能指针的常量向量？

Question

我有一个叫 Thing 的 class。我制作了一个 shared_ptr<Thing>> 的向量。现在我想将它传递给一个函数或类似这样的东西：const vector<shared_ptr<const Thing>>.

下面的代码可以编译，但为了类型安全，我宁愿避免使用 reinterpret_cast，因为我不知道这会如何影响智能指针的行为。

#include <memory>
#include <vector>

using namespace std;

class Thing {};

typedef vector<shared_ptr<Thing>>             VectorOfThings;
typedef const vector<shared_ptr<const Thing>> VectorOfConstThings;

int main() {
    VectorOfThings *things;
    auto           constThings = reinterpret_cast<VectorOfConstThings *> (things);
}

Answer 1

这与智能指针无关； C++ 标准不允许在 container 和 container 之间进行转换。这个之前有讲过： Why is a vector of pointers not castable to a const vector of const pointers?

很可能将 conainter 转换为 container 是未定义的行为。

Answer 2

我写了一个宏来处理类型转换。希望这将消除因编写所有样板代码而产生的错误。

TypeHelper.h

#ifndef TYPE_HELPER_H
#define TYPE_HELPER_H

#include <memory>

/**
 * Creates a struct that has two typedefs and has several functions for reinterpret casting the regular type as
 * the constant type
 * @param NAME The name of the generated struct
 * @param REGULAR_TYPE The typedef of the non-constant type
 * @param CONSTANT_TYPE The typedef of the constant type
 * @param ... The template parameters used in REGULAR_TYPE and CONSTANT_TYPE.
 * There must be at least one parameter passed in. Even if the typedefs don't actually use them.
 * TODO If the typedefs don't need a template parameter, the user should not have to supply a dummy one.
 * TODO Implement some kind of checking to make sure that constness is the only difference.
 * TODO what happens if the template is specialized for a const of that type?
 */
#define TYPE_HELPER( NAME, REGULAR_TYPE, CONSTANT_TYPE, ... ) \
template< __VA_ARGS__ > \
struct NAME { \
typedef REGULAR_TYPE regular; \
typedef CONSTANT_TYPE constant; \
static inline constant *cast( regular *pointer ) { return reinterpret_cast<constant *> (pointer); } \
static inline constant &cast( regular &reference ) { return *reinterpret_cast<constant *> (&reference); }  \
static inline std::unique_ptr<constant> cast( std::unique_ptr<regular> unique_ptr ) { \
    return std::move( *( reinterpret_cast<std::unique_ptr<constant> *> (&unique_ptr))); \
} \
static inline std::shared_ptr<constant> cast( std::shared_ptr<regular> shared_ptr ) { \
    return std::move( *( reinterpret_cast<std::shared_ptr<constant> *> (&shared_ptr))); \
} \
};


#endif //TYPE_HELPER_H

在此文件中，我使用宏定义了 stl 向量的包装器 class

#ifndef VECTOR_TYPE_H
#define VECTOR_TYPE_H

#include <vector>
#include "TypeHelper.h"

TYPE_HELPER( VectorType,
             std::vector<std::shared_ptr<T>>,
             const std::vector<std::shared_ptr<const T>>,
             typename T );

#endif //VECTOR_TYPE_H

最后，我在这里使用类型

#include "VectorType.h"

//This is a vector of std::shared_ptr<std::string>
VectorType<std::string>::regular strings;

//This is a const vector of std::shared_ptr<const str::string>
//note that I needed to use the cast function to assign the regular one to the constant one
VectorType<std::string>::constant constStrings = VectorType<std::string>::cast(strings);

// You don't have to actually type it twice. Use auto.
auto autoConstStrings = VectorType<std::string>::cast(strings);

如何将智能指针向量转换为包含指向常量的智能指针的常量向量？

How do you cast a vector of smart pointers to a constant vector that hold smart pointers that point to constants?

c++

containers

casting

smart-pointers