指向数组返回类型的C ++指针

Question

I have a code like that:

#include <iostream>
using std::cout; 

const int ARR_SIZE=5;
int arr[ARR_SIZE];

int (*five(int first))[ARR_SIZE]
{
    int result[ARR_SIZE];
    for (int i=0;i!=5;++i)
    {
        result[i]=(i+first);
    }
    decltype(result) *final_arr=&result;
    return final_arr;
}

template <typename T>
void print_arr(T *beg, T *end)
{
    cout << "[";
    for (;beg!=end;++beg)
        cout << *beg << ", ";
    cout << "]" << endl;
}

int main()
{
    decltype(arr) *a;
    for (int i=1;i!=5;++i)
    {
        a=five(i);
        print_arr(std::begin(*a),std::end(*a));
    }

    return 0;
}

Basically I have a function that returns pointer to array and I would like to print the contents of this array. I would expect this code to print four arrays:

[1,2,3,4,5]  
[2,3,4,5,6]  
[3,4,5,6,7]  
[4,5,6,7,8]  

However content of the printed arrays seems random. I would be greatful for a hint on what is wrong with the code.

Answer 1

You are going into undefined behaviour. You return a dangling pointer in five .

int result[ARR_SIZE]; life-time is limited by scope of five , therefore it gets freed at the end. Taking its address and returning it does not prolong its life.

This decltype(result) *final_arr=&result; effectively doesn't do anything except hiding the error. You could have skipped it and write return &result; which would most probably upset compiler seriously enough to not compile the code.

Answer 2

您的代码会产生未定义的行为-从five函数返回悬空指针-这意味着您将返回一个指向对象的指针，该对象已分配在堆栈上。