功能对象 [英] Function objects
问题描述
考虑以下代码，
#include< iostream>
struct Functor {
void operator（）（void）{std :: cout<< "函子" <<的std :: ENDL; }
};
int
main（无效）
{
Functor（）;
返回0;
}
为什么运营商（）不需要是否创建了临时对象？
-
anderberg：人们只是突变。有些比其他人差。
* anderberg：请考虑以下代码，
#include< iostream>
struct Functor {
void operator（）（void）{std :: cout<< "函子" <<的std :: ENDL; }
''void''正式参数列表是C-ism;不要。
此外，应该是''const''。
};
int
main（void）
''void''正式参数列表是C-ism;不要。
{
Functor（）;
返回0;
''main''中没有必要，因为''main''是一项特殊功能。
}
为什么操作符（）没有调用'创建的临时对象？
为什么要这样？
尝试Functor（）（）。
-
答：因为它弄乱了人们通常阅读文本的顺序。 />
问：为什么这么糟糕？
A：热门发布。
问：usenet上最烦人的事情是什么？电子邮件？
anderberg写道：考虑以下代码，
＃include< iostream>
struct Functor {
void operator（）（void）{std :: cout<< "函子" <<的std :: ENDL; }
};
int
main（无效）
{
Functor（）;
返回0;
}
为什么operator（）没有调用''创建的临时对象？
因为你实际上并没有*打电话给它。
Functor（）
只是创建一个临时对象。现在，添加（）到它，你得到
Functor（）（）
将导致_call_。
然后，从括号内删除那些''void'。他们就像疮。
唉！
V
2005-12- 07，Alf P. Steinbach写道：* anderberg：考虑以下代码，
#include< iostream>
struct Functor {
void operator（）（void）{std :: cout<< "函子" <<的std :: ENDL; }
''void''正式参数列表是C-ism;不要。
喜欢明白。
另外，应该是''const''。
当然，这不是关于代码''正确性'的问题。
只是一个简单的例子来说明我的问题。< br>{
Functor（）;
返回0;
没有必要''main''，因为''main''是一个特殊功能。
见上文。
为什么不调用operator（）来创建'？blockquote>
为什么要这样？
Functor a;
a（）;
我认为Functor（）会转换成类似于
$的东西b $ b以上。我想我错了。但问题仍然存在 - 为什么
不是吗？
尝试Functor（）（）。
谢谢！
-
anderberg：人们只是突变。有些人比其他人差。
Consider the following code,
#include <iostream>
struct Functor {
void operator()(void) { std::cout << "Functor" << std::endl; }
};
int
main(void)
{
Functor();
return 0;
}
Why isn''t operator() called for the temporary object that''s created?
--
anderberg: People are just mutations. Some worse than others.解决方案* anderberg:Consider the following code,
#include <iostream>
struct Functor {
void operator()(void) { std::cout << "Functor" << std::endl; }
''void'' as formal argument list is C-ism; don''t.
Also, should probably be ''const''.
};
int
main(void)
''void'' as formal argument list is C-ism; don''t.
{
Functor();
return 0;
Not necessary in ''main'', because ''main'' is a special function.
}
Why isn''t operator() called for the temporary object that''s created?
Why should it?
Try Functor()().
--
A: Because it messes up the order in which people normally read text.
Q: Why is it such a bad thing?
A: Top-posting.
Q: What is the most annoying thing on usenet and in e-mail?
anderberg wrote:Consider the following code,
#include <iostream>
struct Functor {
void operator()(void) { std::cout << "Functor" << std::endl; }
};
int
main(void)
{
Functor();
return 0;
}
Why isn''t operator() called for the temporary object that''s created?
Because you didn''t actually *call* it.
Functor()
is just creation of a temporary object. Now, add () to it and you get
Functor()()
which will result in a _call_.
And, drop those ''void'' from inside the parentheses. They''re like sores.
Ugh!
V
On 2005-12-07, Alf P. Steinbach wrote:* anderberg:Consider the following code,
#include <iostream>
struct Functor {
void operator()(void) { std::cout << "Functor" << std::endl; }
''void'' as formal argument list is C-ism; don''t.
Like to be explicit.
Also, should probably be ''const''.
Of course, this wasn''t a question about the code''s "correctness".
Just a simple case of illustrating my question.
{
Functor();
return 0;
Not necessary in ''main'', because ''main'' is a special function.
See above.
Why isn''t operator() called for the temporary object that''s created?
Why should it?
Functor a;
a();
I figured that Functor() would translate to something similar to
the above. I guess I was wrong. But the question remains - why
doesn''t it?
Try Functor()().
Thanks!
--
anderberg: People are just mutations. Some worse than others.
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