[英]Use of overloaded operator ambiguous
如下代码:
typedef void HELPER;
const HELPER* helper = _helper;
inline ostream& operator <<(ostream& out, const HELPER* arg)
{ out << (const char*)(arg); return out; }
如果我尝试炸毁
cout << helper;
具体来说,我得到:
main.cpp:35:28:错误:重载运算符'<<'的用法含糊(操作数类型为'basic_ostream>'和'const HELPER *'(aka'const void *'))
它列出了一些候选人:
/Library/Developer/CommandLineTools/usr/bin/../include/c++/v1/ostream:207:0: note: candidate function
basic_ostream& operator<<(const void* __p);
^
main.cpp:25:17: note: candidate function
inline ostream& operator <<(ostream& out, const HELPER* arg)
^
/Library/Developer/CommandLineTools/usr/bin/../include/c++/v1/ostream:195:20: note: candidate function
basic_ostream& operator<<(bool __n);
^
我的typedef没有在这里调用更强的类型匹配,我感到有些惊讶。 如何使此运算符过载运行?
编辑:进一步澄清,此代码的目的是我双重目标是一组Arduino库。 他们经常通过以下方式管理字符串:
typedef void __FlashStringHelper;
void showHelp(const __FlashStringHelper* helpText)
{
Serial.print(helpText);
}
我喜欢iostream并计划在此双重目标上使用,因此我在Serial对象上重载了<<,并将先前的对象放入了其中(例如,这是过分简化的版本)
#define cout Serial
void showHelp(const __FlashStringHelper* helpText)
{
cout << helpText;
}
现在,我实际上想将真正的iostream定位为其他拱门,但是旧的Arduino代码与__FlashStringHelpers不能有太大的不同。 那是我在的地方
typedef
不会创建类型,而是将其作为别名,
inline ostream& operator <<(ostream& out, const HELPER* arg)
相当于
inline ostream& operator <<(ostream& out, const void* arg)
也许您想创建一个名为HELPER的类型
class HELPER{};
当Zekian回答您的问题时,这可能对您有用或可以帮助您实现您想要做的事情。
#include <iostream>
template <class T>
class Helper {
private:
T obj_;
public:
explicit Helper<T>( T obj ) : obj_(obj) {}
public:
T getObj() const { return obj_; }
void setObj( T obj ) { obj_ = obj; }
template<class U>
inline friend std::ostream& operator<< ( std::ostream& out, const Helper<U>& rhs );
};
template<class U>
std::ostream& operator<< ( std::ostream& out, const Helper<U>& rhs ) {
return out << rhs.obj_;
}
int main() {
Helper<int> helper( 3 );
std::cout << helper << std::endl;
return 0;
}
它是包装类模板,带有重载的ostream运算符<<。 这将适用于整数和原子类型。 如果传递另一个结构或类对象,则必须为其定义其他重载的ostream运算符。
示例 -相同的类模板,但是这次使用类或结构。
#include <iostream>
template <class T>
class Helper {
private:
T obj_;
public:
explicit Helper<T>( T obj ) : obj_(obj) {}
public:
T getObj() const { return obj_; }
void setObj( T obj ) { obj_ = obj; }
template<class U>
inline friend std::ostream& operator<< ( std::ostream& out, const Helper<U>& rhs );
};
template<class U>
std::ostream& operator<< ( std::ostream& out, const Helper<U>& rhs ) {
return out << rhs.obj_;
}
struct Staff {
int employees = 4; // Default to 4
};
int main() {
Staff staff;
Helper<Staff> helper( staff );
std::cout << helper << std::endl; // will not compile
return 0;
}
要修复此ostream,需要为Staff对象重载运算符
template <class T>
class Helper {
private:
T obj_;
public:
explicit Helper<T>( T obj ) : obj_(obj) {}
public:
T getObj() const { return obj_; }
void setObj( T obj ) { obj_ = obj; }
template<class U>
inline friend std::ostream& operator<< ( std::ostream& out, const Helper<U>& rhs );
};
template<class U>
std::ostream& operator<< ( std::ostream& out, const Helper<U>& rhs ) {
return out << rhs.obj_;
}
struct Staff {
int employees = 4;
inline friend std::ostream& operator<< ( std::ostream& out, const Staff& rhs );
};
std::ostream& operator<<( std::ostream& out, const Staff& rhs ) {
return out << rhs.employees;
}
int main() {
Staff staff;
Helper<Staff> helper( staff ); // Default to 4
std::cout << helper << std::endl; // Will Print 4
// To Change Staff's Employee count for the helper wrapper do this:
staff.employees = 12; // Change To 12
helper.setObj( staff ); // pass the changed struct back into helper
std::cout << helper3 << std::endl; // Will Now Print 12
// And For Other Default Types
Helper<int> helper2( 3 );
std::cout << helper2 << std::endl;
Helper<float> helper3( 2.4f );
std::cout << helper3 << std::endl;
return 0;
}
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.