Can I std::move() an element out of a std::vector? [duplicate]

Question

I have a std::vector<std::string> to be re-used in a loop. Is it ok to std::move elements out? If I moved the ith element out, then ith slot goes into an undefined but valid state, but what about the vector? Is its state still defined and valid? Also, can I clear() and then reuse the vector in the next iteration?

EDIT: please read the question before you flag for duplication. I'm asking the state of v2 after doing std::move(v2[0]), not std::move(v2). Also I reuse v2 after I did v2.clear(). How is that similar to the suggested duplication?

EDIT: code example:

struct Foo {     string data;     /* other data memebers */     void workOnData(); }  std::vector<std::string> buffer; Foo foo; while (1) {     buffer.clear();     loadData(buffer); // push data to buffer, at least one element in buffer guaranteed     foo.data.assign(std::move(buffer[0])); // please don't ask is Foo necessary or why workOnData has to be a member method. THIS IS A SIMPLIFIED EXAMPLE!     foo.workOnData(); } 

Answer 1

Is it ok to std::move elements out?

Yes, it's okay.

If I moved the ith element out, then ith slot goes into an undefined but valid state

It leaves the element in a valid but unspecified state. The difference between those two words is important in C++, but probably not too important to this question. Just thought I should point it out.

but what about the vector?

The vector is in a valid and well-specified state, though one of its elements is in an unspecified state.

Also, can I clear() and then reuse the vector in the next iteration?

Yes. Of course. Though you can do a lot more than that. Unlike a moved-from vector, you can still count on the vector having the same size, the same capacity, and all elements other than the one you just moved from remaining unchanged. As well as all references and iterators to elements (including the one you just moved from) remaining valid.