初始化向量< string>双花括号 [英] Initializing vector<string> with double curly braces

问题描述

在下面的示例中，有人可以解释用双花括号和单花括号初始化之间的行为差​​异吗？

Can someone explain the difference in behavior between initializing with double and single curly braces in the example below?

代码＃1：

vector<string> v = {"a", "b"};
string c(v[0] + v[1]);
cout << "c = " << c;
cout << "c.c_str() = " << c.c_str();

输出＃1：

c = ab
c.c_str() = ab

---

代码＃2：

vector<string> v = {{"a", "b"}};
string c(v[0] + v[1]);
cout << "c = " << c;
cout << "c.c_str() = " << c.c_str();

输出2：

c = a\acke�Z\ 
c.c_str() = a

推荐答案

隐式转换中心。就是这样。

Implicit conversion central. That's what's going on.

1. vector< string> v = { a， b}; 通过提供带有两个元素的初始化列表来初始化向量。从字符串文字中初始化两个 std :: string s，然后将其复制到向量中。

1. vector<string> v = {"a", "b"}; You initialize the vector by providing an initializer list with two elements. The two std::strings are initialized from the string literals, and then copied into the vector.

vector< string> v = {{ a， b}}; 通过为初始化器提供 one 元素来初始化向量。单个 std :: string 是从具有两个元素的初始化程序初始化的。访问向量的第二个元素具有不确定的行为。

vector<string> v = {{"a", "b"}}; You initialize the vector by providing an initializer with one element. And that single std::string is initialized from an initializer that has two elements. Accessing the second element of the vector has undefined behavior.

现在，这是有趣的部分。甚至在您访问 v [1] 之前，您的第二个代码段都具有未定义的行为。重载解析（构造单个 std :: string ）选择一个构造函数。最好的可行方法是：

Now, here's the fun part. Your second snippet has undefined behavior even before you access v[1]. Overload resolution (to construct the single std::string) picks a constructor. The best viable one, is this:

template< class InputIt >
basic_string( InputIt first, InputIt last, 
              const Allocator& alloc = Allocator() );

将 InputIt 推导出为 char const [2] （和功能参数调整，将其更改为 char const * ）。既然这些并不是真正的迭代器，那么一切都将变得一团糟。

With InputIt being deduced as char const [2] (and function parameter adjustment, turning it to char const*). Since those aren't really iterators, all hell breaks loose.

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