等号是否在大括号初始化中有所不同？ [英] Does the equals sign make a difference in brace initialization?

问题描述

这里有两种方法来初始化C ++ 11中的变量：

  T a {something} 
 T a = {something}; 
  

我测试了这两个在我能想到的情况，我没有注意到一个区别。 此回答表明两者之间存在细微差别：

<对于变量，我不太注意 T t = {init}; 或 T t { init}; styles，我发现它的区别是小的，并且在最坏的情况下，只会导致一个有用的编译器消息，关于一个显式的构造函数。

$ b

解决方案

我知道的唯一重大差异是在处理显式构造函数：

  struct foo 
 {
 explicit foo（int）; 
}; 
 
 foo f0 {42}; // OK 
 foo f1 = {42}; //不允许
  

这与传统初始化类似：

  foo f0（42）; // OK 
 foo f1 = 42; //不允许
  

请参阅[over.match.list] / 1。

---

除此之外，还有一个缺陷（请参阅 T a = {something} CWG 1270 ） >

  struct aggr 
 {
 int arr [5]; 
}; 
 
 aggr a0 = {1,2,3,4,5}; // OK 
 aggr a1 {1,2,3,4,5}; //不允许
  

Here are two ways to initialize a variable in C++11:

T a {something};
T a = {something};

I tested these two in all scenarios I could think of and I failed to notice a difference. This answer suggests that there is a subtle difference between the two:

For variables I don't pay much attention between the T t = { init }; or T t { init }; styles, I find the difference to be minor and will at worst only result in a helpful compiler message about misusing an explicit constructor.

So, is there any difference between these two?

解决方案

The only significant difference I know is in the treatment of explicit constructors:

struct foo
{
    explicit foo(int);
};

foo f0 {42};    // OK
foo f1 = {42};  // not allowed

This is similar to the "traditional" initialization:

foo f0 (42);  // OK
foo f1 = 42;  // not allowed

See [over.match.list]/1.

---

Apart from that, there's a defect (see CWG 1270) in C++11 that allows brace-elision only for the form T a = {something}

struct aggr
{
    int arr[5];
};

aggr a0 = {1,2,3,4,5};  // OK
aggr a1 {1,2,3,4,5};    // not allowed

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