Using unique_ptr and vector

Question

I have this code:

v8::Handle<v8::Value> StartMethod(const v8::Arguments &args) {
    v8::HandleScope scope; // node_isolate
    int length = args.Length();
    std::vector<std::unique_ptr<char[]>> argv;
    for(int i=0;i<length;++i) {
        if(args[i]->IsString()) {
            v8::String::Utf8Value str(args[i]);
            const int strLen = ToCStringLen(str);
            if(strLen) {
                std::unique_ptr<char []> data(new char[strLen+1]);
                strcpy_s(data.get(), strLen+1, ToCString(str));
                argv.push_back(std::move(data));
            }
        }
    }
    return scope.Close(v8::Int32::New(MainMethod(argv.size(), &(argv[0]._Myptr))));
}

I am using std::move and it is working fine. When i do not use std::move it gives me compiler errors because of unavailability of assignment function.

But Does it guarantees that when vector changes its location, may be because of some internal re sizing and move objects around, nothing bad would happen ?

Am I the only one that noticed the &(argv[0]._Myptr) in the last line of this? Accessing that pointer through this mechanism is bad enough, I don't even want to guess why you're passing the address of it to... something. You would be hard-pressed to find something less portable or standard. — WhozCraig
– WhozCraig, Commented Sep 20, 2013 at 7:47
@WhozCraig I actually wanted char ** at that location. Can you suggest a better method there. — Ashish Negi
– Ashish Negi, Commented Sep 20, 2013 at 9:36
Yes. char *tmp = data.get();, and use &tmp I assume this is a strtol-type function, and if so, and it returns termination info in the resulting pointer-to-pointer (i.e. it modified the pointer) the behavior is beyond undefined. Even accessing that member is implementaiton-specific, Use a temp ptr anyway. What you're doing now with &(argv[0]._Myptr) is dreadful. — WhozCraig
– WhozCraig, Commented Sep 20, 2013 at 14:16

Community · Accepted Answer · 2017-05-23 11:56:35Z

2

vector<T>::push_back gives the strong exception safety guarantee for move-only types if and only if the move construtor of T does not throw.

In your case T is a unique_ptr for array objects, whose move constructor is declared noexcept (§20.7.1.3), so you are indeed fine here: If the internal resizing of the vector throws a bad_alloc, the vector will remain unchanged and usable.

edited May 23, 2017 at 11:56

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answered Sep 20, 2013 at 7:51

ComicSansMS

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Comments

Theolodis · Accepted Answer · 2013-09-20 07:13:14Z

1

I am not sure if I got your question right, but under the assumption that std::vector doesn't have any unexpected behaviour, yes, it is garantueed. By executing std::move() you transfer the control over the underlaying pointer to the std::unique_ptr in the std::vector. From this point on it should behave like a std::vector<char*>

answered Sep 20, 2013 at 7:13

Theolodis

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3 Comments

Ashish Negi Over a year ago

Would it still retain properties of unique_ptr like deleting after going out of scope ?

us2012 Over a year ago

Yes (in the sense that your allocated data is deleted when the vector argv goes out of scope - and in the sense that an individual piece of data is deleted when its unique_ptr is erased from the vector).

Theolodis Over a year ago

of course, executing std::move() transfers the control of the underlaying object to the std::unique_ptr<char*> in the std::vector<std::unique_ptr<char*>>, what that means is, that it is like if the old std::unique_ptr<char*> had never existed.

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