c++ - T t {x} 和 T t = { x } 对于整数或枚举类型的区别？

Question

在 C++14 中:

对于任何整数或枚举类型 T 以及对于任何表达式 expr:

有没有区别:

struct S { T t { expr }; };

和

struct S { T t = { expr }; };

更新:

我得到了 [dcl.init.list]p3b5 上面写着:

If the initializer list has a single element of type E and either T is not a reference type or its referenced type is reference-related to E, the object or reference is initialized from that element.

我相信这句话适用于direct-list-initialization 和copy-list-initialization。

所以我认为答案是否定的，没有区别。

Answer 1

如果你看一下direct initialization和 copy initialization引用文献，你会发现相同的词:

if T is a non-class type, standard conversions are used, if necessary, to convert the value of other to the cv-unqualified version of T

所以应该没有区别。这些初始化的区别仅适用于类类型:复制初始化不考虑显式构造函数和显式用户-定义的转换运算符，直接初始化。整型和枚举类型都没有。

编辑:
@ᐅ Johannes Schaub - litb ᐊ answered a relative question to this one (仅关于括号，而不是大括号)并且他引用了 8.5/14 类似的措辞(强调我的):

The form of initialization (using parentheses or =) is generally insignificant, but does matter when the initializer or the entity being initialized has a class type; see below. If the entity being initialized does not have class type, the expression-list in a parenthesized initializer shall be a single expression.

我也无法在标准中找到对应的 {}。我希望这些论据足以支持没有区别的答案。