Is initializing with “var{args}” a new feature of

问题:

I was reading the C++0x faq and came across the section detailing initializer lists. The examples were mostly variations of:

vector<int> vi = { 1, 2, 3 };
vector<int> vj({1, 2, 3});
// etc.

However, also listed was the form:

vector<int> vk{2};

This form appears elsewhere in the faq, and I am curious as to whether it is semantically different from the initial two forms, or just syntactic sugar for vk({x, y, z}).

回答1:

One is uniform initialization, and the other is initializer lists. They are two different things, although as you can see, they can produce similar syntax.

vector<int> vk{2};

is a uniform initialization- the other two are initializer lists.

回答2:

The ({1, 2, 3}) form calls the constructors of vector<int> directly, and passes as first argument a {1, 2, 3}. You could have passed more arguments

vector<int> vk({1, 2, 3}, myAllocator);

If vector<int> would not have a constructor whose first parameter is an initializer_list or of another type that could be initialized by {1, 2, 3} (like, another container class), it would not work. In your case it works because vector<int> in fact has a constructor whose first parameter is a initializer_list<int>. This is just like in normal function calls

void f(vector<int> const& vk);
int main() { f({1, 2, 3}); }

If you omit the parentheses, as in vector<int> vk{1, 2, 3}, the exact meaning depends on the class. A vector<int> has an initializer list constructor, which is a constructor with a first parameter of type initializer_list<int> (optionally a reference to it), and all other params with default arguments. If the class has such a constructor, then the initializer list is passed to that constructor. Alternatively the class could simply be an aggregate (like struct A { int a; int b; int c; };, the initializer list would then init the members) or have a constructor that accepts 3 separate int arguments.

Finally the = { 1, 2, 3 } form is almost identical to the version omitting the parentheses (i.e just removing =), except that it forbids to use explicit constructors (i.e had they declared it as explicit vector(initializer_list<int>); or had they declared a explicit vector(int, int, int); instead, it would result in an error if you use = { 1, 2, 3 }).

回答3:

The uniform initialization prevents narrowing conversions i.e. conversions that would cause loss of data:

#include <vector>

std::vector<float> v{1.0F, 2.0F, 3.0F}; // OK: 

std::vector<float> w{1.0, 2.0, 3.0}; // OK: doubles could be put into floats without loss.

std::vector<int> j{1.1, 2.2, 3.3}; // error: narrowing

std::vector<int> k{1L, 2L, 3L}; // OK: the long numbers can be represented as int without loss.

std::vector<int> l{0xfacebeefL, 0xdeadbabeL, 0xfadecabeL}; // error: narrowing.