在C ++ 11中'typedef'和'using'有什么区别? [英] What is the difference between 'typedef' and 'using' in C++11?
问题描述
我知道在C ++ 11中,我们现在可以使用使用
写入类型别名,例如 typedef
:
I know that in C++11 we can now use using
to write type alias, like typedef
s:
typedef int MyInt;
根据我的理解,等同于:
Is, from what I understand, equivalent to:
using MyInt = int;
新的语法出现在努力表达 template typedef
:
And that new syntax emerged from the effort to have a way to express "template typedef
":
template< class T > using MyType = AnotherType< T, MyAllocatorType >;
但是,前两个非模板示例,标准中是否有其他细微差别?例如, typedef
s以弱的方式进行别名操作。这是它不创建一个新的类型,但只有一个新的名称(转换是隐含在这些名称之间)。
But, with the first two non-template examples, are there any other subtle differences in the standard? For example, typedef
s do aliasing in a "weak" way. That is it does not create a new type but only a new name (conversions are implicit between those names).
与使用相同
还是生成新类型?
推荐答案
它们是相同的,从标准(强调我的)(7.1.3.2):
They are equivalent, from the standard (emphasis mine) (7.1.3.2):
typedef名称也可以通过别名声明引入。 using关键字之后的
标识符成为typedef名称,并且标识符后面的
可选属性 - 说明符seq在该typedef名称之后为
。 它具有与由typedef说明符引入的
相同的语义。特别地,它
不定义一个新类型,它不会出现在type-id 。
A typedef-name can also be introduced by an alias-declaration. The identifier following the using keyword becomes a typedef-name and the optional attribute-specifier-seq following the identifier appertains to that typedef-name. It has the same semantics as if it were introduced by the typedef specifier. In particular, it does not define a new type and it shall not appear in the type-id.
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