How to construct a std::string from a std::vector?

Question

I\'d like to build a std::string from a std::vector.

I could use std::stringsteam, but imagine there is a s

Answer 1

C++03

std::string s;
for (std::vector::const_iterator i = v.begin(); i != v.end(); ++i)
    s += *i;
return s;

C++11 (the MSVC 2010 subset)

std::string s;
std::for_each(v.begin(), v.end(), [&](const std::string &piece){ s += piece; });
return s;

C++11

std::string s;
for (const auto &piece : v) s += piece;
return s;

Don't use std::accumulate for string concatenation, it is a classic Schlemiel the Painter's algorithm, even worse than the usual example using strcat in C. Without C++11 move semantics, it incurs two unnecessary copies of the accumulator for each element of the vector. Even with move semantics, it still incurs one unnecessary copy of the accumulator for each element.

The three examples above are O(n).

std::accumulate is O(n²) for strings.

You could make std::accumulate O(n) for strings by supplying a custom functor:

std::string s = std::accumulate(v.begin(), v.end(), std::string{},
    [](std::string &s, const std::string &piece) -> decltype(auto) { return s += piece; });

Note that s must be a reference to non-const, the lambda return type must be a reference (hence decltype(auto)), and the body must use += not +.

C++20

In the current draft of what is expected to become C++20, the definition of std::accumulate has been altered to use std::move when appending to the accumulator, so from C++20 onwards, accumulate will be O(n) for strings, and can be used as a one-liner:

std::string s = std::accumulate(v.begin(), v.end(), std::string{});