“默认构造”内置类型？ [英] "Default construction" of built-in types?
问题描述
我最近在模板中遇到了一些代码，默认构造了一个类型为T的
对象传递给另一个函数...
SomeFunction（ T（））;
实例化该模板的代码将T指定为int。
正确的程序行为依赖于默认构造" int为零。
引导我进入下面的例子......
int main（）
{
int i; //未初始化！预计自动内置类型
int j = int（）; //初始化为零！不确定会发生什么
返回0;
}
这样的默认构造是什么？内置类型标准C ++？
char（）== 0
int（）== 0
等...
谢谢
" DaKoadMunky" <哒********* @ aol.com>写了...我最近在一个模板中遇到了一些代码，默认构造了一个类型为T的
对象传递给另一个函数...
SomeFunction （T（））;
实例化该模板的代码将T指定为int。
正确的程序行为依赖于默认构造。 int是
零。
" Default-initialised"是正确的术语。并且，是的，它应该是
初始化为零。
这引导我进入以下示例...
int main（ ）
{
int i; //未初始化！预计自动内置式
int j = int（）; //初始化为零！不确定会发生什么
评论的第二部分是什么意思？
返回0;
}
<这种默认结构是什么？内置类型标准C ++？
是的。
Victor
>> int j = int（）; //初始化为零！不确定会发生什么
评论的第二部分是什么意思？
评论本来可以延长在其他编译器，平台等上说
，因为我不知道零的初始化是否是标准的。
Victor Bazarov写道：" DaKoadMunky" <哒********* @ aol.com>写道...
我最近在一个模板中遇到了一些代码，默认构造了一个类型为T的
对象传递给另一个函数...
SomeFunction（T（））;
实例化该模板的代码将T指定为int。
正确的程序行为依赖于默认构造。 int is
零。
"默认初始化"是正确的术语。并且，是的，它应该被初始化为零。
这引导我进入以下示例...
int main（）
{i /; //未初始化！预计自动内置式
int j = int（）; //初始化为零！不确定会发生什么
评论的第二部分是什么意思？
return 0;
}
这样的默认构造是什么？内置类型标准C ++？
是的。
Victor
我不相信你，所以我查了一下，果然，这是正确的。
到OP：
标准说初始化程序为（）的任何对象是
" value-initialized"，对于POD类型，value-initialized"定义
为零初始化。有关详细信息，请参阅第8.5节。
Alan
I recently came across some code in a template that default constructed an
object of type T to pass to another function...
SomeFunction(T());
The code that instantiates that template specifies T as an int.
Proper program behavior relies on that "default constructed" int being zero.
That lead me to the following example...
int main()
{
int i; //Uninitialized! Expected for automatic of built-in type
int j = int(); //Initialized to zero! Not sure what to expect
return 0;
}
Is such "default construction" of built-in types standard C++?
char() == 0
int()==0
etc...
Thanks
"DaKoadMunky" <da*********@aol.com> wrote...I recently came across some code in a template that default constructed an
object of type T to pass to another function...
SomeFunction(T());
The code that instantiates that template specifies T as an int.
Proper program behavior relies on that "default constructed" int being zero.
"Default-initialised" is the proper term. And, yes, it is supposed to be
initialised to zero.
That lead me to the following example...
int main()
{
int i; //Uninitialized! Expected for automatic of built-in type
int j = int(); //Initialized to zero! Not sure what to expect
What do you mean by the second part of the comment?
return 0;
}
Is such "default construction" of built-in types standard C++?
Yes.
Victor
>> int j = int(); //Initialized to zero! Not sure what to expect
What do you mean by the second part of the comment?
Comment could have been extended to say "on other compilers, platforms, etc.."
given that I didn''t know if the initialization to zero was standard.
Victor Bazarov wrote:"DaKoadMunky" <da*********@aol.com> wrote...I recently came across some code in a template that default constructed an
object of type T to pass to another function...
SomeFunction(T());
The code that instantiates that template specifies T as an int.
Proper program behavior relies on that "default constructed" int being
zero.
"Default-initialised" is the proper term. And, yes, it is supposed to be
initialised to zero.That lead me to the following example...
int main()
{
int i; //Uninitialized! Expected for automatic of built-in type
int j = int(); //Initialized to zero! Not sure what to expect
What do you mean by the second part of the comment?return 0;
}
Is such "default construction" of built-in types standard C++?
Yes.
Victor
I didn''t believe you, so I looked it up, and sure enough, that is correct.
To the OP:
The standard says that any object whose initializer is () is
"value-initialized", and for POD types, "value-initialized" is defined
as being "zero-initialized". Refer to Section 8.5 for more info.
Alan
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