C++: creating an array of local objects

Question

If I do the following:

string* create_array(){
    string* arr = new string[2];
    string s = "hello";
    string s2 = "world";

    arr[0] = s;
    arr[1] = s2;
    return arr;
}

is the assignment of arr[0] = s making a copy of s and then putting that copy in the memory address that arr[0] points to? or is it making arr[0] refer to the local stack variable s, and in this case if I use the array returned from this function calling arr[0] will have unpredictable behaviour?

Thanks

It is a bad practice to return an array allocated with new from a function. Consider using std::vector or std::array instead. You usually should not mess with raw pointers when you don't sure you need to, or, especially, if you doubt in how they work. — Mikhail
– Mikhail, Commented Nov 13, 2012 at 14:20
@Mikhail I am using arrays over vectors for speed as I am doing some real time image processing stuff. — Aly
– Aly, Commented Nov 13, 2012 at 14:22
@Aly there is no speed difference, std::array should actually be faster. — Šimon Tóth
– Šimon Tóth, Commented Nov 13, 2012 at 14:25
@Let_Me_Be I am definitely seeing a tangible speed difference (albeit microseconds) between raw pointer and vector — Aly
– Aly, Commented Nov 13, 2012 at 14:26

Luchian Grigore · Accepted Answer · 2012-11-13 14:17:48Z

4

is the assignment of arr[0] = s making a copy of s

Yes.

and then putting that copy in the memory address that arr[0] points to?

No. arr[0] is a string, it doesn't point to anything. (internally, it probably has a char* somewhere, but that's unique per instance of std::string and is an implementation detail, I added this comment just to be complete)

or is it making arr[0] refer to the local stack variable s, and in this case if I use the array returned from this function calling arr[0] will have unpredictable behaviour?

Nope, it makes a copy.

answered Nov 13, 2012 at 14:17

Luchian Grigore

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