为什么向量返回会错过一些值？

Question

int main(int argc, char argv)
{
    int myarray[] = {1, 2, 3, 5};
    std::vector<int> array(myarray, myarray + 4);
    std::vector<int> *p = testvector(array);
    std::vector<int>::const_iterator it;
    for(it=p->begin(); it != p->end(); ++ it)
    {
        printf("%d ", *it);
    }
    return 0;
}

std::vector<int> *testvector(std::vector<int> array) 
{
    return &array;
}

Above is my test code; what is wrong that it returned 0 0 3 5 instead of 1 2 3 5

Answer 1

Look at std::vector<int> *testvector(std::vector<int> array) carefully. This is taking a deep copy of the input parameter array .

The returned pointer will be dangling once array is out of scope. The behaviour on dereferencing that pointer will be undefined . That's why your program is behaving oddly.

Interestingly, if you had written std::vector<int>& array as the parameter (ie passed by reference ) then this would have worked, since you would be returning a pointer to the vector defined in main ! This code would be extremely brittle though.

Answer 2

this:

std::vector<int> *testvector(std::vector<int> array) 

should be replaced with:

std::vector<int> *testvector(std::vector<int>& array) 

Live Demo

However, this is not a real solution since there is no point from your example. I think you have an X problem and you are just posting the Y one. if you post your real problem, it will be easier to help you.

Answer 3

Change

std::vector<int> *testvector(std::vector<int> array) 

to

std::vector<int> *testvector(std::vector<int>& array) // Note the &

Otherwise the array is copied to a local variable, which is destroyed at the end of testvector() which invalidates the returned pointer.