C++ 使用结构（无类）cout 样式重载“<<”
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overloading "<<" with a struct (no class) cout style
提问by monkeyking
I have a struct that I'd like to output using either 'std::cout' or some other output stream. Is this possible without using classes?
我有一个结构，我想使用“std::cout”或其他一些输出流来输出。这可能不使用类吗？
Thanks
谢谢
#include <iostream>
#include <fstream>
template <typename T>
struct point{
T x;
T y;
};
template <typename T>
std::ostream& dump(std::ostream &o,point<T> p) const{
o<<"x: " << p.x <<"\ty: " << p.y <<std::endl;
}
template<typename T>
std::ostream& operator << (std::ostream &o,const point<T> &a){
return dump(o,a);
}
int main(){
point<double> p;
p.x=0.1;
p.y=0.3;
dump(std::cout,p);
std::cout << p ;//how?
return 0;
}
I tried different syntax' but I cant seem to make it work.
我尝试了不同的语法'但我似乎无法让它工作。
回答by GManNickG
Perhaps it's a copy-paste error, but there are just a few things wrong. Firstly, free-functions cannot be const, yet you have marked dumpas such. The second error is that dumpdoes not return a value, which is also easily remedied. Fix those and it should work:
也许这是一个复制粘贴错误，但只是有一些错误。首先，自由函数不能是const，但你已经标记dump为这样。第二个错误是dump不返回值，这也很容易修复。修复这些，它应该可以工作：
template <typename T> // note, might as well take p as const-reference
std::ostream& dump(std::ostream &o, const point<T>& p)
{
return o << "x: " << p.x << "\ty: " << p.y << std::endl;
}
回答by Joris Timmermans
For all intents and purposes, structs areclasses in C++, except that their members default to public instead of private. There are potentially minor implementation-specific differences because of optimization but these have no effect on the standard functionality which is the same for classes and structs in C++.
出于所有意图和目的，结构是C++中的类，除了它们的成员默认为 public 而不是 private。由于优化，可能存在细微的特定于实现的差异，但这些对标准功能没有影响，这与 C++ 中的类和结构相同。
Secondly, why have the "dump" function? Just implement it directly in the stream operator:
其次，为什么要有“转储”功能？直接在流操作符中实现即可：
template<typename T>
std::ostream& operator << (std::ostream& o, const point<T>& a)
{
o << "x: " << a.x << "\ty: " << a.y << std::endl;
return o;
}
