上QQ阅读APP看书，第一时间看更新

How it works...

Without auto type deduction, there's not much to be surprised about in the brace {} operator, at least, when initializing regular types. When initializing containers such as std::vector, std::list, and so on, a brace initializer will match the std::initializer_list constructor of that container class. It does this in a greedy manner, which means that it is not possible to match non-aggregate constructors (non-aggregate constructors are usual constructors in contrast to the ones that accept an initializer list).

std::vector, for example, provides a specific non-aggregate constructor, which fills arbitrarily many items with the same value: std::vector<int> v (N, value). When writing std::vector<int> v {N, value}, the initializer_list constructor is chosen, which will initialize the vector with two items: N and value. This is a special pitfall one should know about.

One nice detail about the {} operator compared to constructor calling using normal () parentheses is that they do no implicit type conversions: int x (1.2); and int x = 1.2; will initialize x to value 1 by silently rounding down the floating point value and converting it to int. int x {1.2};, in contrast, will not compile because it wants to exactly match the constructor type.

One can controversially argue about which initialization style is the best one.
Fans of the bracket initialization style say that using brackets makes it very explicit, that the variable is initialized with a constructor call, and that this code line is not reinitializing anything. Furthermore, using {} brackets will select the only matching constructor, while initializer lines using () parentheses try to match the closest constructor and even do type conversion in order to match.

The additional rule introduced in C++17 affects the initialization with auto type deduction--while C++11 would correctly make the type of the variable auto x {123}; an std::initializer_list<int> with only one element, this is seldom what we would want. C++17 would make the same variable an int.

Rule of thumb:

auto var_name {one_element}; deduces var_name to be of the same type as one_element
auto var_name {element1, element2, ...}; is invalid and does not compile
auto var_name = {element1, element2, ...}; deduces to an std::initializer_list<T> with T being of the same type as all the elements in the list

C++17 has made it harder to accidentally define an initializer list.

Trying this out with different compilers in C++11/C++14 mode will show that some compilers actually deduce auto x {123}; to an int, while others deduce it to std::initializer_list<int>. Writing code like this can lead to problems regarding portability!