C ++ 11：string（50，'x'）与string {50，'x'} [英] C++11: string(50, 'x') versus string{50, 'x'}

问题描述

如 ideone 所示：

  cout<< string（50，'x'）; // xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx 
 cout<< string {50，'x'}; // 2x 
  

WAT ??

我发现50是ASCII'2'，所以：

  cout< static_cast< int>（'2'）; // 50 
 cout<< static_cast< char>（50）; // 2 
  

但是我已经有了。

解决方案

当你做

code> string {50，'x'} 你基本上用字符列表初始化字符串。

另一方面， string（50，'x'）调用一个2参数构造函数，它被定义为重复 x 50次。 string {50，'x'} 不选择构造函数的原因是它可能是不明确的。如果你有一个三参数构造函数呢？如果类型有一个 initializer_list 构造函数，当你使用 {...} 进行初始化时， / p>

基本上你需要知道你的类型的构造函数。 initializer_list 构造函数总是有优先级，以避免歧义。

As seen on ideone:

cout << string(50, 'x'); // xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
cout << string{50, 'x'}; // 2x

WAT??

I have figured out that 50 is ASCII '2', so:

cout << static_cast<int>('2'); // 50
cout << static_cast<char>(50); // 2

But that's as far as I've got.

Does this lead to a solid argument against C++11 initializers?

解决方案

When you do string { 50, 'x' } you're essentially initializing the string with a list of characters.

On the other hand, string(50, 'x') calls a 2 argument constructor, which is defined to repeat the character x 50 times. The reason why string { 50, 'x' } doesn't pick the constructor is that it could be ambiguous. What if you had a three parameter constructor as well? If the type has an initializer_list constructor, it will be picked when you use { ... } for initialization.

Basically you need to be aware of the constructors your type has. The initializer_list constructor will always have a precedence to avoid ambiguity.

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