英文:
Curly Brace Initialisation with std::string
问题
我最近越来越多地使用花括号初始化。尽管我在这种情况下发现与圆括号初始化有所不同，我想知道为什么。
如果我这样做：
```c++
const std::string s(5, '=');
std::cout << s << std::endl;
我会得到：
=====
这是我预期的结果。但是如果我这样做：
const std::string s{5, '='};
std::cout << s << std::endl;
我会得到：
=
为什么会这样？
编辑：为了让看到这个的任何人受益，第二个输出中的 `=` 前面有一个不可打印的字符。它在stackoverflow上不显示。看起来像这样：
英文:
I have been using curly brace initialisation more and more recently. Although I found a difference to round bracket initialisation in this case and I was wondering why.
If I do:
const std::string s(5, '=');
std::cout << s << std::endl;
I get:
=====
This is what I expect. But if I do:
const std::string s{5, '='};
std::cout << s << std::endl;
I get:
=
Why is this?
Edit: For the benefit for anyone that sees this. There is an unprintable character before the = in the second output. It just doesn't show on stackoverflow. Looks like:
答案1
得分: 5
这部分内容的翻译如下：
const std::string s(5, '=');
使用了接受_count_和字符_ch_（以及分配器）的构造函数：
constexpr basic_string(size_type count, CharT ch,
const Allocator& alloc = Allocator());
但是
const std::string s{5, '='};
使用了
constexpr basic_string(std::initializer_list<CharT> ilist,
const Allocator& alloc = Allocator());
这意味着5将被转换为一个char，因此你的字符串的大小将是2。第一个字符将具有值5，而另一个字符将是=。
英文:
This
const std::string s(5, '=');
uses the constructor taking a count and the character, ch (as well as an allocator):
constexpr basic_string( size_type count, CharT ch,
const Allocator& alloc = Allocator() );
But
const std::string s{5, '='};
uses
constexpr basic_string( std::initializer_list<CharT> ilist,
const Allocator& alloc = Allocator() );
which means that 5 will be converted to a char and your string will therefore have the size 2. The first character will have the value 5 and the other will be =.
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