C++: std::vector::resize vs. “normal” allocation

Question

In the code example for std::transform , there is an example with code like this:

std::vector<int> foo;
std::vector<int> bar;

//add some elements to foo

bar.resize(foo.size());

//store elements transformed from foo's in bar

And I was wondering whether

std::vector<int> bar;    
bar.resize(foo.size());

was any different from

std::vector<int> bar(foo.size());

and if so, how?

Answer 1

没有什么区别，但后者是一_^点点更高效和简洁。

Answer 2

No, there's no difference. At least not in the way you show it (with no insertions into foo between the definition of bar and the call to resize ).