通过默认构造函数语义进行初始化 [英] Iinitialisation via default constructor semantics
问题描述
我对另一个关于此问题的帖子感到困惑：
这种初始化是这样的吗？
int x = int（）;
初始化为0的类型仅适用于内置类型或*所有* POD类型？
即：
struct any
{
int x，y;
} a = whatever（）;
是保证ax和ay初始化为0？
-
Ioannis Vranos
http://www23.brinkster.com/noicys
I was confused from another thread about this:
Does this kind of initialisation
int x= int();
initialise to 0 of the type only for built in types or for *all* POD types?
That is:
struct whatever
{
int x,y;
}a=whatever();
is guaranteed a.x and a.y to be initialised to 0?
--
Ioannis Vranos
http://www23.brinkster.com/noicys
推荐答案
" Ioannis Vranos" < iv*@guesswh.at.grad.com>在消息新闻中写道：ci ********** @ ulysses.noc.ntua.gr ...
"Ioannis Vranos" <iv*@guesswh.at.grad.com> wrote in message news:ci**********@ulysses.noc.ntua.gr...
仅为内置类型初始化为0的类型或* *所有* POD类型？
initialise to 0 of the type only for built in types or for *all* POD types?
T x = T（）;
复制使用默认初始化对象初始化x，无论类型T是什么，
内置，POD或其他。
T x = T();
copy initializes x with a Default-Initialized object regardless of what the type T is,
builtin, POD, or otherwise.
Ron Natalie写道：
Ron Natalie wrote:
Ioannis Vranos < iv*@guesswh.at.grad.com>在消息新闻中写道：ci ********** @ ulysses.noc.ntua.gr ...
"Ioannis Vranos" <iv*@guesswh.at.grad.com> wrote in message news:ci**********@ulysses.noc.ntua.gr...
仅初始化为0的类型对于内置类型或* all * POD类型？
initialise to 0 of the type only for built in types or for *all* POD types?
T x = T（）;
复制使用默认初始化对象初始化x，无论什么是T类型，内置，POD或其他。
T x = T();
copy initializes x with a Default-Initialized object regardless of what the type T is,
builtin, POD, or otherwise.
对于内置类型，这相当于0对应的类型。 />
对于非POD类型，结果与T x（T（））相同;
对于除内置类型之外的POD类型，是否与内置<的情况相同类型为
？
TC ++ PL仅提供内置类型，用于第一种初始化。
-
Ioannis Vranos
http：// www23 .brinkster.com / noicys
" Ioannis Vranos" < iv*@guesswh.at.grad.com>在消息中写道
新闻：ci ********** @ ulysses.noc.ntua.gr ...
"Ioannis Vranos" <iv*@guesswh.at.grad.com> wrote in message
news:ci**********@ulysses.noc.ntua.gr...
这样吗初始化
int x = int（）;
初始化为0类型仅适用于内置类型或*所有* POD
类型？
Does this kind of initialisation int x= int(); initialise to 0 of the type only for built in types or for *all* POD
types?
所有POD类型。
此外，在C ++ 2003（但不是C ++ 1998）中，它适用于非POD
结构的POD成员，前提是它们没有明确说明定义的构造函数。
示例：
struct Foo {
int x;
std :: string s;
};
struct Bar {
int x;
std :: string s;
Bar（）{}
};
在C ++ 1998中，Foo（ ）.x未定义;在C ++ 2003中，它是0.
在C ++ 1998和C ++ 2003中，Bar（）。x都是未定义的。原因是Bar的
作者有机会初始化x并选择不这样做，
而Foo的作者没有机会初始化x ，
定义了没有提供此类机会的构造函数。
All POD types.
Moreover, in C++2003 (but not C++1998), it works for POD members of non-POD
structures, provided that they have no explicitly defined constructors.
Example:
struct Foo {
int x;
std::string s;
};
struct Bar {
int x;
std::string s;
Bar() { }
};
In C++1998, Foo().x is undefined; in C++ 2003, it''s 0.
In both C++1998 and C++2003, Bar().x is undefined. The reason is that the
author of Bar had the opportunity to initialize x and chose not to do so,
whereas the author of Foo did not have the opportunity to initialize x,
having defined no constructor that would have provided such an opportunity.
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