诚心求教
程序如下：
#include <iostream>
#include <string>
using namespace std;
class Teacher
{
public:
Teacher(string nam,int a,string t)
{name=nam;
age=a;
title=t;
}
void display()
{
cout < < "name: " < <name < <endl;
cout < < "age: " < <age < <endl;
cout < < "title: " < <title < <endl;
}
protected:
string name;
int age;
string title;
};
class Student
{public:
Student(char nam[],char s,float sco)
{
strcpy(name1,nam);
sex=s;
score=sco;
}
void display1()
{ cout < < "name: " < <name1 < <endl;
cout < < "sex: " < <sex < <endl;
cout < < "score: " < <score < <endl;
}
protected:
string name1;
char sex;
float score;
};
class Graduate:public Teacher,public Student
{
public:
Graduate(string nam,int a,char s,string t,float sco,float w):
Teacher(nam,a,t),Student(nam,s,sco),wage(w){}
void show()
{cout < < "name: " < <name < <endl;
cout < < "age: " < <age < <endl;
cout < < "sex: " < <sex < <endl;
cout < < "score: " < <score < <endl;
cout < < "title: " < <title < <endl;
cout < < "wages: " < <wage < <endl;
}
private:
float wage;
};
int main()
{
Graduate grad1( "wangli ",25, 'f ', "asssdfdf ",96.1,2546.3);
grad1.show();
return 0;
}
用visual c++编译提示：
error C2664: 'strcpy ' : cannot convert parameter 1 from 'class std::basic_string <char,struct std::char_traits <char> ,class std::allocator <char> > ' to 'char * '
No user-defined-conversion operator available that can perform this conversion, or the operator cannot be called
不知道是什么意思。。。。
[解决办法]
strcpy(name1,nam);
改为：
name1 = nam;
[解决办法]
Student 类中 strcpy(name1,nam); 改成 name1 = nam;
更推荐的做法是把这个放在成员初始化列表里初始化.
[解决办法]
意思就是你调用strcpy的参数不对
strcpy的原型为char *strcpy(char *str1, const char *str2);
你的name1是一个string无法转换成char *，你可以使用name1=string(nam);实现你想要的功能
------解决方案--------------------
建议你编程时随手有一本标准库的参考比如C&C++.Programmers.Reference
到网上可以搜索到电子版
[解决办法]
在student类中，先把char nam[]改为string nam，strcpy(name1,nam)改为name1=nam
构造函数是用来初始化的，即付值，strcpy(name1,nam)虽然能起到相同的作用，但意义不同