为什么使用auto的直接列表初始化被认为是不好的或不受欢迎的?

问题描述

我已经习惯使用如下所示的直接列表初始化来编写代码，因为它更有效，而且对于防止隐式缩小:

I've come into the habit of writing code with direct-list-initialization like below as it's more effective and it's very useful to prevent implicit narrowing:

int i {0};
string s {""};
char c {'a'};
bool b {false};

auto num {100}; // But this??

但是当涉及到自动说明符时，我听说这样写它被认为是不好的或者不是首选，那为什么呢?

But when it comes to the auto specifier, I have heard it is considered bad or not preferred to write it like that, why is that?

推荐答案

以下是使用该语法失败的示例:

Here's an example of where using that syntax fails:

struct Foo{};

void eatFoo (const Foo& f){}

int main() {
    Foo a;
    auto b{a};
    eatFoo(b);
}

您可能希望这没问题:b应该是Foo，并传递给eatFoo.不幸的是，这导致以下编译器错误:

You might expect this to be fine: b should be a Foo and be passed to eatFoo. Unfortunately, this results in the following compiler error:

prog.cpp:11:10: error: invalid initialization of reference of type 'const Foo&' from expression of type 'std::initializer_list<Foo>'
  eatFoo(b);

如您所见，b实际上是std::initializer_list<Foo>类型.在这种情况下，当然不是我们想要的.如果我们将其更改为auto b = a，则可以正常工作.然后，如果我们仍然想使用auto，但要明确声明类型，则可以将其更改为auto b = Foo{a}，然后让编译器删除该副本.

As you can see, b is actually of type std::initializer_list<Foo>. Certainly not what we want in this case. If we change it to auto b = a, this works fine. Then if we want to still use auto, but explicitly state the type, we can change it to auto b = Foo{a} and let the compiler elide the copy.

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