C ++做一个POD typedef的值初始化? [英] Does C++ do value initialization of a POD typedef?
问题描述
C ++在简单的 POD typedefs上执行值初始化吗?
Does C++ do value initialization on simple POD typedefs?
假设
typedef T* Ptr;
Ptr()
做值初始化并保证等于 (T *)0
?
do value-initialization and guarantee to equal (T*)0
?
例如
Ptr p = Ptr();
return Ptr();
推荐答案
对于 T
, T()
值初始化类型 T
并生成一个右值表达式。
It does. For a type T
, T()
value-initializes an "object" of type T
and yields an rvalue expression.
int a = int();
assert(a == 0);
与pod-classes相同:
Same for pod-classes:
struct A { int a; };
assert(A().a == 0);
对于没有用户声明的构造函数的一些非POD类也是如此:
Also true for some non-POD classes that have no user declared constructor:
struct A { ~A() { } int a; };
assert(A().a == 0);
由于您不能 A a()
(创建一个函数声明),boost有一个类 value_initialized
,允许解决此问题,C ++ 1x将具有以下替代语法
Since you cannot do A a()
(creates a function declaration instead), boost has a class value_initialized
, allowing to work around that, and C++1x will have the following, alternative, syntax
int a{};
在标准的干燥语言中,这听起来像
In the dry words of the Standard, this sounds like
表达式T(),其中T是非数组完整对象类型或(可能是cv限定的)void类型的简单类型说明符(7.1.5.2) ,创建一个指定类型的右值,它是值初始化的
The expression T(), where T is a simple-type-specifier (7.1.5.2) for a non-array complete object type or the (possibly cv-qualified) void type, creates an rvalue of the specified type, which is value-initialized
由于typedef名是一个类型名,一个简单类型说明符本身,这工作很好。
Since a typedef-name is a type-name, which is a simple-type-specifier itself, this works just fine.
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