返回 stl 容器的正确方法

Question

A lot of times, we need to return a container constructed inside a function. Say I have a function to get all primes in some range.

std::vector<int> getPrimes() {
    std::vector<int> primes;
    ...
    reutrn primes;
}

This seems need to copy all primes to the returned object. Then pointers are needed.

// =================Wrong code=================
// std::vector<int> *getPrimes() {
//     std::vector<int> primesptr;
//     ...
//     reutrn &primesptr;
// }

But apparently local variables cannot be returned. So the new operator can be used.

std::vector<int> *getPrimes() {
    std::vector<int> *primesptr = new std::vector<int>();
    ...
    reutrn primesptr;
}

Finally, this meets the need. But it's ugly and complicated (because of the usage of pointers). So what's the recommended way to return STL containers constructed inside a function? Moreover, if I return the STL container by value, will it be optimized so that only a constant amount of values or pointers are copied (instead of copying every element in the container)?

Answer 1

If you're using a modern compiler, just use your original code; NRVO (Named Return Value Optimization), a form of copy elision , is performed by most modern compilers in the simple case you're demonstrating.

If for some reason you can't rely on this, and profiling shows this is a big problem on this crappy compiler you're using, you can, if you insist, explicitly do:

std::vector<int> getPrimes() {
    std::vector<int> primes;
    ...
    return std::move(primes);
}

This will prevent NRVO (so you'll definitely end up move-constructing the return value while destructing the emptied local variable), but move-construction of a vector is a fixed cost unrelated to the vector 's size (a handful of pointer and size_t assignments, the main backing array's storage is untouched).