Are there any cases where it is incorrect to replace push_back with emplace_back?

Question

Can I break a valid C++03 program by replacing std::vector::push_back with emplace_back and compiling it with C++ 11 compiler? From reading e

Answer 1

I constructed a short example that actually fails to compile when push_back is replaced by emplace_back:

#include <vector>
struct S {
    S(double) {}
  private:
    explicit S(int) {}
};
int main() {
    std::vector<S>().push_back(0); // OK
    std::vector<S>().emplace_back(0); // error!
}

The call to push_back needs to convert its argument 0 from type int to type S. Since this is an implicit conversion, the explicit constructor S::S(int) is not considered, and S::S(double) is called. On the other hand, emplace_back performs direct initialization, so both S::S(double) and S::S(int) are considered. The latter is a better match, but it's private, so the program is ill-formed.

Answer 2

Yes, you can change the behavior (more than just avoiding a copy constructor call), since emplace_back only sees imperfectly forwarded arguments.

#include <iostream>
#include <vector>
using namespace std;

struct Arg { Arg( int ) {} };

struct S
{
    S( Arg ) { cout << "S(int)" << endl; }
    S( void* ) { cout << "S(void*)" << endl; }
};

auto main()
    -> int
{
    vector<S>().ADD( 0 );
}

Example builds:

[H:\dev\test\0011]
> g++ foo.cpp -D ADD=emplace_back && a
S(int)

[H:\dev\test\0011]
> g++ foo.cpp -D ADD=push_back && a
S(void*)

[H:\dev\test\0011]
> _

---

Addendum: as pointed out by Brian Bi in his answer, another difference that can lead to different behavior is that a push_back call involves an implicit conversion to T, which disregards explicit constructors and conversion operators, while emplace_back uses direct initialization, which does consider also explicit constructors and conversion operators.

Answer 3

Suppose a user-defined class could be initialized from braced-initializer. e.g.

struct S {
    int value;
};

then

std::vector<S> v;

v.push_back({0});    // fine
v.emplace_back({0}); // template type deduction fails

std::vector::emplace_back is template function but std::vector::push_back is non-template function. With a braced-initializer std::vector::emplace_back would fail because template argument deduction fails.

Non-deduced contexts

6) The parameter P, whose A is a braced-init-list, but P is not std::initializer_list or a reference to one:

LIVE

Answer 4

The emplace versions don't create an object of the desired type at all under exception circumstances. This can lead to a bug.

Consider the following example, which uses std::vector for simplicity (assume uptr behaves like std::unique_ptr, except the constructor is not explicit):

std::vector<uptr<T>> vec;
vec.push_back(new T());

It is exception-safe. A temporary uptr<T> is created to pass to push_back, which is moved into the vector. If reallocation of the vector fails, the allocated T is still owned by a smart pointer which correctly deletes it.

Compare to:

std::vector<uptr<T>> vec;
vec.emplace_back(new T());

emplace_back is not allowed to create a temporary object. The ptr will be created once, in-place in the vector. If reallocation fails, there is no location for in-place creation, and no smart pointer will ever be created. The T will be leaked.

Of course, the best alternative is:

std::vector<std::unique_ptr<T>> vec;
vec.push_back(make_unique<T>());

which is equivalent to the first, but makes the smart pointer creation explicit.

Answer 5

If you don't have crazy side-effects in copy constructor of the objects that you hold in your vector, then no.

emplace_back was introduced to optimise-out unnecessary copying and moving.

Answer 6

int main() {
std::vector<S>().push_back(0);
std::vector<S>().emplace_back(0); 
}

Give struct constructor in emplace_back i.e The above code will be like this

int main() {
std::vector<S>().push_back(0); 
std::vector<S>().emplace_back(S(0)); 
}