我正在尝试重载ostream和istream作为模板中类的freind。我已经在线查看,但是找不到关于模板重载的更多内容,我所看到的似乎也表明这是一个重要的方式来实现这些。很明显,我对编程非常陌生,并希望得到任何帮助。谢谢。
#include <stdio.h>
#include<vector>
#include<iostream>
using namespace std;
template<class T>
class MyClass
{
enter code here
public:
MyClass();
MyClass(const T& p_val1);
MyClass(const MyClass<T>& p_val1);
~MyClass();
MyClass<T>& operator=(MyClass<T>& rhs);
friend ostream& operator<<(ostream& lhs, const MyClass<T> &printme);
friend istream& operator>><T>(istream& lhs, MyClass<T>& readme);
private:
T* m_val1;
};
实现ostream和istream。
template<class T>
ostream& operator<<(ostream&lhs, const MyClass<T>& printme)
{
lhs << printme.m_val1;
return lhs;
}
template<class T>
istream& operator>>(istream& lhs, MyClass<T>& readme)
{
lhs >> *(readme.m_val1);
return lhs;
}
这是错误
Undefined symbols for architecture x86_64:
"MyClass<int>::~MyClass()", referenced from:
_main in main.o
"operator<<(std::__1::basic_ostream<char, std::__1::char_traits<char> >&, MyClass<int> const&)", referenced from:
_main in main.o
ld: symbol(s) not found for architecture x86_64
clang: error: linker command failed with exit code 1 (use -v to see invocation)
主要功能
MyClass<int> a(8);
MyClass<int> b(9);
cout << " Enter a value for the two types to be swapped: ";
cin >> a >> b;
cout << "Before swapping...\n";
cout << "Object a's pointer member dereferences a value of:\t" << a << endl;
cout << "Object b's pointer member dereferences a value of:\t" << b << endl;
答案 0 :(得分:0)
您正在声明您的运营商,但没有定义它们。模板定义与类中的模板定义无关,因为它们的原型不同,因此链接错误。将您的声明更改为:
template <typename T, typename Trait>
friend std::basic_ostream<T, Trait>&
operator<< (std::basic_ostream<T, Trait>& out, const MyClass& c)
{
return out << c.my_val1;
}
template <typename T, typename Trait>
friend std::basic_istream<T, Trait>&
operator>> (std::basic_istream<T, Trait>& in, MyClass& c)
{
return in >> *(c.my_val1);
}
请注意,它使用std::basic_ostream
和std::basic_istream
及其各自的类型和类型特征。它允许使用所有流,而不仅仅是std::ostream
和std::istream
。