类指针数组浅拷贝：如何修正Picture_Pair的拷贝构造函数？

Let's break down what's wrong with your current code and fix it step by step.

1. Name Collision Problem #

Your copy constructor parameter is named pp, which clashes with your class's private member array pp. When you write pp = pp.pp, you're actually trying to access the parameter's non-existent pp member instead of the source object's pointer array. We need to rename the parameter to avoid this conflict—other is a standard, readable choice.

2. Array Assignment Limitation #

You can't directly assign one array to another using the = operator in C++. For a shallow copy of your pointer array, you need to copy each element individually.

Here's the corrected version of your class with a proper shallow copy constructor:

#ifndef PICTURE_PAIR_H
#define PICTURE_PAIR_H

class Picture; // Forward declaration

class Picture_Pair {
private:
    Picture* pp[2] {nullptr, nullptr}; // Pointers to the two Pictures

public:
    // Shallow copy constructor
    Picture_Pair(const Picture_Pair& other) {
        // Copy each pointer individually (shallow copy)
        pp[0] = other.pp[0];
        pp[1] = other.pp[1];
    }

    // Add a default constructor (required since we defined a user-defined copy constructor)
    Picture_Pair() = default;

    // Don't forget to declare/destructors if you manage Picture object lifetimes!
};

#endif // PICTURE_PAIR_H

What This Does #

• The shallow copy constructor takes a const Picture_Pair& other (the source object we're copying from).
• We copy each pointer in other.pp to our own pp array. This means both the original Picture_Pair and the new copy will point to the same Picture objects—this is exactly the behavior of a shallow copy.

Critical Note #

Since this is a shallow copy, watch out for double deletion if your Picture_Pair class is responsible for deleting the Picture objects. If both the original and copied Picture_Pair try to delete the same pointers in their destructors, you'll get undefined behavior. If you ever need a deep copy (where each Picture_Pair owns its own Picture instances), you'd have to allocate new Picture objects and copy their data instead of just copying pointers.

内容的提问来源于stack exchange，提问作者Micka1895