c++ - 这个 std::vector 构造函数中发生了什么？

Question

我看到一个函数引用了 std::vector，传递给它的参数让我对发生的事情感到困惑。它看起来像这样:

void aFunction(const std::vector<int>& arg) { }


int main()
{
    aFunction({ 5, 6, 4 }); // Curly brace initialisation? Converting constructor?

    std::vector<int> arr({ 5, 6, 4 }); // Also here, I can't understand which of the constructors it's calling

    return 0;
}

谢谢。

Answer 1

对于通过这种结构创建的对象，您需要提供接受 std::initializer_list 的构造函数和 std::vector 有 one (8) :

vector( std::initializer_list<T> init, 
        const Allocator& alloc = Allocator() );

你也可以在那个页面上看到一个例子:

// c++11 initializer list syntax:
std::vector<std::string> words1 {"the", "frogurt", "is", "also", "cursed"}; 

注意:C++11 也允许用花括号初始化对象:

Someobject {
   Someobject( int ){}
};

Someobject obj1(1); // usual way
Someobject obj2{1}; // same thing since C++11

但是你需要小心，如果对象之前提到了 ctor，它将被使用:

std::vector<int> v1( 2 ); // creates vector with 2 ints value 0
std::vector<int> v2{ 2 }; // creates vector with 1 int value 2

注意 2:对于您的问题，列表是如何创建的，在文档中有描述:

A std::initializer_list object is automatically constructed when:

a braced-init-list is used in list-initialization, including function-call list initialization and assignment expressions

a braced-init-list is bound to auto, including in a ranged for loop