类型转换 [英] type conversion

问题描述

参见下面的示例：为什么a的输出工作和

''b''的输出无法编译？有没有办法写类'某事''，以便

''b''默认正确转换？ （包括iostream，字符串，使用

命名空间std）

模板< typename T>

struct something {

T x;

运营商T（）;

};


int

main（ int，char **）

{

something< int> a;

something< string> b;


ax = 4711;

bx =" hello world！" ;;


cout< ;< a<< ENDL; //编译好！

cout<< b<< ENDL; //运算符<< ......没有匹配，编译失败！

}

编译器= gcc 3.3.5


感谢您的帮助。


弗兰克

See example below: Why does the output of ''a'' work and the output of
''b'' fails to compile? Is there a way to write class ''something'' so that
''b'' converts correctly by default? (include iostream, string, use
namespace std)
template <typename T>
struct something {
T x;
operator T();
};

int
main(int, char**)
{
something<int> a;
something<string> b;

a.x = 4711;
b.x = "hello world!";

cout << a << endl; // compiles fine!
cout << b << endl; // operator<< ... no match, compile fails!
}
compiler = gcc 3.3.5

Thanks for any Help.

Frank

推荐答案

frs写道：

参见示例下面：为什么a的输出工作和
b的输出无法编译？有没有办法写类'某事''，以便
''b''默认正确转换？ （包括iostream，string，use
namespace std）

template< typename T>
struct something {
T x;
operator T（） ;
};

int
main（int，char **）
{
某事< int> a;
某事< string> b;

a.x = 4711;
b.x =" hello world！" ;;

cout<< a<< ENDL; //编译好！
cout<< b<< ENDL; //运算符<< ......没有匹配，编译失败！

编译器= gcc 3.3.5

感谢您的帮助。

Frank

See example below: Why does the output of ''a'' work and the output of
''b'' fails to compile? Is there a way to write class ''something'' so that
''b'' converts correctly by default? (include iostream, string, use
namespace std)
template <typename T>
struct something {
T x;
operator T();
};

int
main(int, char**)
{
something<int> a;
something<string> b;

a.x = 4711;
b.x = "hello world!";

cout << a << endl; // compiles fine!
cout << b << endl; // operator<< ... no match, compile fails!
}
compiler = gcc 3.3.5

Thanks for any Help.

Frank

这可能与字符串本身是一个

模板有关。

例如以下编译


模板< typename T>

struct something {

T x ;

运算符T（）;

};


struct S

{

};


ostream& operator<<（ostream&，const S&）;


int main（int，char **）

{

某事< int> a;

某事< S> b;


a.x = 4711;

b.x = S（）;


cout<< a<< ENDL; //编译好！

cout<< b<< ENDL; //汇编很好！

}


我不知道你问题的完整答案，但我的建议真的很好

会远离这些东西。我认为即使是BS也说要远离语言中那些不起眼的角落。


John

It''s probably got something to do with the fact that string itself is a
template.

For instance the following does compile

template <typename T>
struct something {
T x;
operator T();
};

struct S
{
};

ostream& operator<<(ostream&, const S&);

int main(int, char**)
{
something<int> a;
something<S> b;

a.x = 4711;
b.x = S();

cout << a << endl; // compiles fine!
cout << b << endl; // compiles fine!
}

I don''t know the complete answer to your question but really my advice
would be to stay away from this stuff. I think even BS said stay away
from the obscure corners of the language.

John

" John Harrison" <乔************* @ hotmail.com>在消息中写道

新闻：TJ ************** @ newsfe7-win.ntli.net ...

"John Harrison" <jo*************@hotmail.com> wrote in message
news:TJ**************@newsfe7-win.ntli.net...

frs写道：

参见下面的例子：为什么a的输出工作和
b的输出无法编译？有没有办法写类'某事''，以便
''b''默认正确转换？ （包括iostream，string，use
namespace std）

template< typename T>
struct something {
T x;
operator T（） ;
};

int
main（int，char **）
{
某事< int> a;
某事< string> b;

a.x = 4711;
b.x =" hello world！" ;;

cout<< a<< ENDL; //编译好！
cout<< b<< ENDL; //运算符<< ......没有匹配，编译失败！

编译器= gcc 3.3.5

感谢您的帮助。

Frank

See example below: Why does the output of ''a'' work and the output of
''b'' fails to compile? Is there a way to write class ''something'' so that
''b'' converts correctly by default? (include iostream, string, use
namespace std)
template <typename T>
struct something {
T x;
operator T();
};

int
main(int, char**)
{
something<int> a;
something<string> b;

a.x = 4711;
b.x = "hello world!";

cout << a << endl; // compiles fine!
cout << b << endl; // operator<< ... no match, compile fails!
}
compiler = gcc 3.3.5

Thanks for any Help.

Frank

这可能与字符串本身是一个
模板有关。

例如以下编译

模板< typename T>
struct something {
T x;
运营商T（）;
};

struct S
{
};

ostream& operator<（&ostream&，const S&）;

int main（int，char **）
{
某事< int> a;
某事< S> b;

a.x = 4711;
b.x = S（）;

cout<< a<< ENDL; //编译好！
cout<< b<< ENDL; //编译好！
}

我不知道你问题的完整答案，但我的建议真的是远离这些东西。我认为即使是BS也要远离语言的模糊角落。

John

It''s probably got something to do with the fact that string itself is a
template.

For instance the following does compile

template <typename T>
struct something {
T x;
operator T();
};

struct S
{
};

ostream& operator<<(ostream&, const S&);

int main(int, char**)
{
something<int> a;
something<S> b;

a.x = 4711;
b.x = S();

cout << a << endl; // compiles fine!
cout << b << endl; // compiles fine!
}

I don''t know the complete answer to your question but really my advice
would be to stay away from this stuff. I think even BS said stay away from
the obscure corners of the language.

John

如果是这样的话那么这条线


something< string> b;


将无法编译。


一种可能性是使用的字符串不是std :: string而且不是' '$

提供带流的输出操作。


Ben

Had that been the case then the line

something<string> b;

would have failed to compile.

One possibility is that the string used wasn''t std::string and doesn''t
provide output operation with streams.

Ben

嗯，它确实没有'用std :: string编译......我无法弄清楚原因是什么'

...


无论如何以下OP会做OP可能想做的事情：


模板< typename T>

struct something

{

T x;

};


模板< typename OStreamT，typename T>

OStreamT&运算符<< （

OStreamT& os，const something< T>& o）

{

return os<< o.x;

}


int main（无效）

{

something< int> a;

something< std :: string> b;


ax = 4711;

bx =" hello world！" ;;


std： ：cout<< a<< std :: endl;

std :: cout<< b<< std :: endl;

}


ben

hmm, it really didn''t compile with std::string...I can''t figure out what''s
the cause...

Anyway the following will do what the OP might want to do:

template <typename T>
struct something
{
T x;
};

template <typename OStreamT, typename T>
OStreamT& operator << (
OStreamT& os, const something<T>& o)
{
return os << o.x;
}

int main(void)
{
something<int> a;
something<std::string> b;

a.x = 4711;
b.x = "hello world!";

std::cout << a << std::endl;
std::cout << b << std::endl;
}

ben

这篇关于类型转换的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！