Why would I ever use push_back instead of emplace_back?

Solution 1:

I have thought about this question quite a bit over the past four years. I have come to the conclusion that most explanations about push_back vs. emplace_back miss the full picture.

Last year, I gave a presentation at C++Now on Type Deduction in C++14. I start talking about push_back vs. emplace_back at 13:49, but there is useful information that provides some supporting evidence prior to that.

The real primary difference has to do with implicit vs. explicit constructors. Consider the case where we have a single argument that we want to pass to push_back or emplace_back.

std::vector<T> v;
v.push_back(x);
v.emplace_back(x);

After your optimizing compiler gets its hands on this, there is no difference between these two statements in terms of generated code. The traditional wisdom is that push_back will construct a temporary object, which will then get moved into v whereas emplace_back will forward the argument along and construct it directly in place with no copies or moves. This may be true based on the code as written in standard libraries, but it makes the mistaken assumption that the optimizing compiler's job is to generate the code you wrote. The optimizing compiler's job is actually to generate the code you would have written if you were an expert on platform-specific optimizations and did not care about maintainability, just performance.

The actual difference between these two statements is that the more powerful emplace_back will call any type of constructor out there, whereas the more cautious push_back will call only constructors that are implicit. Implicit constructors are supposed to be safe. If you can implicitly construct a U from a T, you are saying that U can hold all of the information in T with no loss. It is safe in pretty much any situation to pass a T and no one will mind if you make it a U instead. A good example of an implicit constructor is the conversion from std::uint32_t to std::uint64_t. A bad example of an implicit conversion is double to std::uint8_t.

We want to be cautious in our programming. We do not want to use powerful features because the more powerful the feature, the easier it is to accidentally do something incorrect or unexpected. If you intend to call explicit constructors, then you need the power of emplace_back. If you want to call only implicit constructors, stick with the safety of push_back.

An example

std::vector<std::unique_ptr<T>> v;
T a;
v.emplace_back(std::addressof(a)); // compiles
v.push_back(std::addressof(a)); // fails to compile

std::unique_ptr<T> has an explicit constructor from T *. Because emplace_back can call explicit constructors, passing a non-owning pointer compiles just fine. However, when v goes out of scope, the destructor will attempt to call delete on that pointer, which was not allocated by new because it is just a stack object. This leads to undefined behavior.

This is not just invented code. This was a real production bug I encountered. The code was std::vector<T *>, but it owned the contents. As part of the migration to C++11, I correctly changed T * to std::unique_ptr<T> to indicate that the vector owned its memory. However, I was basing these changes off my understanding in 2012, during which I thought "emplace_back does everything push_back can do and more, so why would I ever use push_back?", so I also changed the push_back to emplace_back.

Had I instead left the code as using the safer push_back, I would have instantly caught this long-standing bug and it would have been viewed as a success of upgrading to C++11. Instead, I masked the bug and didn't find it until months later.

Solution 2:

push_back always allows the use of uniform initialization, which I'm very fond of. For instance:

struct aggregate {
    int foo;
    int bar;
};

std::vector<aggregate> v;
v.push_back({ 42, 121 });

On the other hand, v.emplace_back({ 42, 121 }); will not work.