[英]Why am I getting this error: nomatch for 'operator<<' in C++?
Please help me out on why I'm getting an error where I have mentioned the comment guys. 请帮我解决为什么我在提到评论员的地方出错了。 I'm really stuck here! 我真的被困在这里! What other details should I give exactly? 我还应该提供哪些其他详细信息?
Main.cpp Main.cpp
#include <iostream>
#include<person.h>
#include <sstream>
int main()
{
cout << "Constructor Overloading Demo !!!" << endl;
person vish;
cout << vish.toString() << endl; /* I get an error for this statement though I feel this is correct */
return 0;
}
person.cpp //constructor definition file person.cpp //构造函数定义文件
#include "person.h"
#include <sstream>
person::person()
{
name = "vish";
age = 25;
//ctor
}
void person::toString()
{
stringstream st;
st << "Name: ";
st << name;
st << "& Age: ";
st << age;
prnt = st.str();
cout << prnt << endl;
//return 10;
}
person.h //constructor declaration person.h //构造函数声明
#ifndef PERSON_H
#define PERSON_H
#include<iostream>
using namespace std;
class person
{
public:
person();
void toString();
//virtual ~person();
protected:
private:
string name;
int age;
string prnt;
};
#endif // PERSON_H
As name of method person::toString()
suggests it should return string
, not void
(which means that method does not return anything): 就像方法person::toString()
建议的那样,它应该返回string
,而不是void
(这意味着该方法不返回任何东西):
class person {
...
std::string toString();
//^--------- type should be changed
of course you need to change that method implementation accordingly: 当然,您需要相应地更改该方法的实现:
std::string person::toString()
{
stringstream st;
st << "Name: ";
st << name;
st << "& Age: ";
st << age;
return st.str();
}
I realize that my function was returning void. 我意识到我的函数正在返回void。 but my question is: why does that matter ? 但是我的问题是:为什么这很重要?
Your output to cout
is this: 您对cout
输出是这样的:
cout << vish.toString(); // endl is omitted for clarity
which is equal to one of: 等于以下之一:
cout.operator<<( vish.toString() );
operator<<( cout, vish.toString() );
whatever is available. 任何可用的。 This means that function or method operator<<
needs result of call to toString()
and it cannot accept void
. 这意味着函数或方法operator<<
需要调用toString()
结果,并且不能接受void
。 If you just want to print from method toString()
write: 如果只想从方法toString()
打印,请编写:
vish.toString();
that would work fine (except method name would be still confusing, but compiler does not care). 可以正常工作(除非方法名仍然令人困惑,但编译器不在乎)。
operator<< expects a string to be passed to it. operator <<期望将一个字符串传递给它。 Your ToString function is returning void 您的ToString函数返回void
string person::toString()
{
stringstream st;
st << "Name: ";
st << name;
st << "& Age: ";
st << age;
return st.str();
}
edit: 编辑:
The statement 该声明
std::cout << vish.toString();
is identical to 与...相同
std::cout.operator<<(vish.toString());
The prototype for the operator is 操作员的原型是
std::ostream& operator<<(std::string);
std::ostream& operator<<(int);
//...
The function expects a string or some other parameter as an argument. 该函数需要一个字符串或其他参数作为参数。 Your function is defined as a standalone print function. 您的功能定义为独立的打印功能。 So it should either be modified as stated above or called by itself. 因此,应按上述说明对其进行修改或对其进行单独调用。
The function name 'toString()' implies that the function returns a string representation of the object that can be used for anything, not just printing, so it should return a string. 函数名称“ toString()”表示该函数返回该对象的字符串表示形式,该对象可以用于任何用途,而不仅仅是打印,因此应返回一个字符串。 If the function prints the object it should be called 'print' instead of 'toString'. 如果函数打印对象,则应将其称为“ print”而不是“ toString”。
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