operator << 操作符重载

[五号智能]

#include <iostream>

struct abc
{	
        int a;
	std::string b;
	double c;


        friend ostream& operator << (ostream &out,abc &hot)
        {
        	out << hot.a << hot.b << hot.c;
	
	        return out;
        }
}

int main()
{
	abc a;	
	 a.a=3;
	 a.b="none";
	 a.c=3.204;
std::cout <<a<<endl;
} 

编译不能功。。我刚学这东西，还不知道哪里错。。帮帮我吧 ：（

[i244782405]

好象是要在头文件下加一句:
using manespace std;吧
std::cout <<a<<endl;
这个是什么哟 我从来没这样用过呢 还请楼下的指教下

[jy01807853]

//帮你改好了GCC上编译通过
#include <iostream>
using namespace std;//引用空间名就不用后面的，std::cout
struct abc
{    
    int a;
    std::string b;
    double c;
    friend std::ostream& operator << (std::ostream &out,abc &hot)
    {
      out << hot.a << hot.b << hot.c;
    
    return out;
    }
};//这点问题
int main()
{
     abc a;    
     a.a=3;
     a.b="none";
     a.c=3.204;
     cout <<a<<endl;
     system("pause");
     return 0;
} 

[五号智能]

啊！！！！！！！！！！！！

[hpj2000]

要细心咯~

[taodm]

如果你打开string这个头文件，那么应该能总结出理由的，如果C++ Primer你认真看了关于函树重载解析的章节

[wkh5432]

呵呵 顶下 学习中

[phoex]

#include <iostream>
using namespace std;
struct abc
{
int a;
std::string b;
double c;


friend ostream& operator << (ostream &out,abc &hot)
{
out << hot.a << hot.b.c_str() << hot.c;

return out;
}
};

void main(void )
{
abc a;
a.a=3;
a.b="none";
a.c=3.204;
std::cout <<a<<endl;
}

[sai_deer]

#include <iostream>
#include <string> //string库头文件~
using namespace std; //对于小程序这样写方便~

struct abc
{
int a;
string b;
double c;

friend ostream& operator << (ostream &out, abc &hot)
{
out << hot.a << hot.b << hot.c;
return out;
}
};

int main()
{
abc a;
a.a = 3;
a.b = "none";
a.c = 3.204;
cout << a << endl;
return 0;
}