c++ - 将字符串的填充构造函数与初始化列表一起使用

Question

我正在通读 Ivor Horton's "Beginning C++" .在第 7 章中，有一个关于字符串初始化的例子:

However, you can initialize a string with any number of instances of given character. You can define and initialize a sleepy time string object like this:

string sleeping {6, 'z'};

The string object, sleeping, will contain "zzzzzz". The string length will be 6.

使用我的编译器 Apple LLVM version 6.0 (clang-600.0.57)，示例无法按描述运行。相反，它就像我给出的那样工作:

 string sleeping {"\6z"};

使用 string's "from c-string" constructor, rather than its "fill" constructor .

我知道我可以通过使用括号来使用填充构造函数:

 string sleeping(6, 'z');

但是为了满足我的好奇心，有没有办法像书中示例那样使用带有初始化列表的填充构造函数？

Answer 1

不，这是不可能的，标准specifies it :

• If T is an aggregate type [...];
• Otherwise, if T is a character array [...];
• If T is an aggregate type [...];
• Otherwise, If the braced-init-list is empty [...];
• Otherwise, the constructors of T are considered, in two phases:
  • All constructors that take std::initializer_list as the only argument, or as the first argument if the remaining arguments have default values, are examined, and matched by overload resolution against a single argument of type std::initializer_list

所以 constructor以 std::initializer_list 作为其第一个参数的 std::string 将始终在接受计数和字符的参数之前被考虑。

在此构造函数中使用列表初始化的唯一方法是同时提供分配器:

std::string s{6, 'z', std::string::allocator_type{}};