运算符<<必须只有一个参数 [英] operator << must take exactly one argument

问题描述

ah

  #includelogic.h
 ... 
 
 A类
 {
 friend ostream& operator<<（ostream& amp; amp;）; 
 ... 
}; 
  

logic.cpp

  #includeah
 ... 
 ostream& logic :: operator<<<（ostream& os，A& a）
 {
 ... 
} 
 ... 
  
 
 
 
当我编译时，它说：
 
 
 ：ostream&逻辑::运算符<<<（std :: ostream& amp;）必须只有一个参数。
 
 
问题？问题是你在类中定义它，其中a）意味着第二个参数是隐式的（<$ p 
 $ 
 this ）和b）它不会做你想做的，即extend  std :: ostream 。您必须将其定义为自由函数：
 类A {/ * ... * /}; 
 std :: ostream& operator<<（std :: ostream& const A& a）; 
  
 
a.h
#include "logic.h"
...

class A
{
friend ostream& operator<<(ostream&, A&);
...
};
logic.cpp
#include "a.h"
...
ostream& logic::operator<<(ostream& os, A& a)
{
...
}
...
When i compile, it says:

  
std::ostream& logic::operator<<(std::ostream&, A&)' must take exactly one argument.
What is the problem?
解决方案
The problem is that you define it inside the class, which a) means the second argument is implicit (this) and b) it will not do what you want it do, namely extend std::ostream. You have to define it as a free function:
class A { /* ... */ };
std::ostream& operator<<(std::ostream&, const A& a);


                        
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