C++11 initializer list fails – but only on lists of length 2

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Introduction

Imagine the following declaration, and usage:

struct A {
  A (std::initializer_list<std::string>);
};


A {{"a"          }}; // (A), initialization of 1 string
A {{"a", "b"     }}; // (B), initialization of 1 string << !!
A {{"a", "b", "c"}}; // (C), initialization of 3 strings

In (A) and (C), each c-style string is causing the initialization of one (1) std::string, but, as you have stated in your question, (B) differs.

The compiler sees that it’s possible to construct a std::string using a begin- and end-iterator, and upon parsing statement (B) it will prefer such construct over using "a" and "b" as individual initializers for two elements.

A { std::string { "a", "b" } }; // the compiler's interpretation of (B)

Note: The type of "a" and "b" is char const[2], a type which can implicitly decay into a char const*, a pointer-type which is suitable to act like an iterator denoting either begin or end when creating a std::string.. but we must be careful: we are causing undefined-behavior since there is no (guaranteed) relation between the two pointers upon invoking said constructor.

Explanation

When you invoke a constructor taking an std::initializer_list using double braces {{ a, b, ... }}, there are two possible interpretations:

  1. The outer braces refer to the constructor itself, the inner braces denotes the elements to take part in the std::initializer_list, or:

  2. The outer braces refer to the std::initializer_list, whereas the inner braces denotes the initialization of an element inside it.

It’s prefered to do 2) whenever that is possible, and since std::string has a constructor taking two iterators, it is the one being called when you have std::vector<std::string> {{ "hello", "there" }}.

Further example:

std::vector<std::string> {{"this", "is"}, {"stackoverflow"}}.size (); // yields 2

Solution

Don’t use double braces for such initialization.

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