C++:防止在 const 函数中更改指针的值

Question

采用这个虚拟结构:

struct foo
{
    int* pI;
    void bar() const
    {
        int* ptrToI = pI; // I want this to require 'const int*'
        *ptrToI = 5; // I want this to fail
    }
}__test__;

如何设计此结构以防止我更改 pI 指向的值？

Answer 1

您可以使用自定义智能指针来隐藏底层指针:

template <typename T>
struct ptr_t
{
private:
    T *ptr;
public:
    //Constructors and assignment operators
    ptr_t(): ptr(nullptr) {}
    ptr_t(T* p): ptr(p) {}
    ptr_t(const ptr_t &other): ptr(other.ptr) {}
    ptr_t& operator=(T* p) {this->ptr=p;return *this;}
    ptr_t& operator=(const ptr_t &other) {this->ptr=other.ptr; return *this;}

    //Note that the smart pointers included in the standard returns non-const pointers
    //That is why you need to define a custom smart pointer, which forces the const pointer return in the const version of the operators.
    const T* operator->() const {return ptr;}
    T* operator->() {return ptr;}
    const T& operator&() const {return *ptr;}
    T& operator&() {return *ptr;}

    operator const T*() const {return ptr;}
    operator T*() {return ptr;}
};

struct foo2
{
    ptr_t<int> pI;
    void nonconst_bar()
    {
        int* ptrToI = pI; // Now success, since not const
        *ptrToI = 5;
    }

    void failing_bar() const
    {
        //int* ptrToI = pI; // This fails
        //*pI = 5; // This also fails
    }

    void success_bar() const
    {
        const int* ptrToI = pI;
        //*ptrToI = 5; // This is not possible anymore
    }
};