c++ - 错误: 'int'之前的预期主表达式以及代码中的许多类似错误

Question

PS:我是初学者。
我尝试的完整代码如下:this_is_the_tree_I_Was_trying_to_code_up
错误出现在第30,50,63,74和82行，例如:

第30行:“int”之前的预期主表达式；“char”之前的预期主表达式；
第50行:“int”之前的预期主表达式；“char”之前的预期主表达式；
第63行:“int”之前的预期主表达式；“char”之前的预期主表达式；
第74行:“int”之前的预期主表达式；“char”之前的预期主表达式；“int”之前的预期主表达式；
第82行:“int”之前的预期主表达式；“char”之前的预期主表达式；“int”之前的预期主表达式；

#include <iostream>
#include <math.h>
#include <cstring>
using namespace std;
class staff{
protected:
    int code;
    char *name;
public:
    staff(){}
    staff(int code1,char *name1){
        code=code1;
        int length;
        length=strlen(name1);
        name=new char[length+1];
        strcpy(name,name1);
    }
    void display(){
        cout<<"Staff code is: "<<code<<"\n";
        cout<<"Name is: "<<name<<"\n";
    }
};
class teacher:public staff{
protected:
    char *subject;
    char *publication;
public:
    teacher(){}
    teacher(int code1,char *name1,char *sub1,char *pub1):
        staff(int code1,char *name1){                      //line 30
        int len1,len2;
        len1=strlen(sub1);
        len2=strlen(pub1);
        subject=new char[len1+1];
        publication=new char[len2+1];
        strcpy(subject,sub1);
        strcpy(publication,pub1);
    }
    void display_t(){
        cout<<"Subject is: "<<subject<<"\n";
        cout<<"Publication is: "<<publication<<"\n";
    }
};
class officer:public staff{
protected:
    char grade;
public:
    officer(){}
    officer(int code1,char *name1,char grd1):
        staff(int code1,char *name1){                       //line 50
        grade=grd1;
    }
    void display_o(){
        cout<<"Grade is: "<<grade;
    }
};
class typist:public staff{
protected:
    int speed;
public:
    typist(){}
    typist(int code1,char *name1,int spd):
        staff(int code1,char *name1){                         //line 63
        speed=spd;
    }
    void display_typ(){
        cout<<"Speed is: "<<speed<<"\n";
    }
};
class regular:public typist{
public:
    regular(){}
    regular(int code1,char *name1,int spd):
        typist(int code1,char *name1,int spd){}            //line 74
};
class casual:public typist{
protected:
    float dailywage;
public:
    casual(){}
    casual(float wage,int code1,char *name1,int spd):
        typist(int code1,char *name1,int spd){             //line 82
        dailywage=wage;
    }
    void display_c(){
        cout<<"Daily wage of casual typists are: "<<dailywage<<"\n";
    }
};
int main()
{
    casual c1;
    c1=casual(2100.05,002,"Ganesh",50);
    c1.display();
    //c1.display_t();
    c1.display_typ();
    //c1.display_o();
    c1.display_c();
    return 0;
}

Answer 1

在这行上:

teacher(int code1,char *name1,char *sub1,char *pub1):
        staff(int code1,char *name1) {

您使用了错误的语法来调用基类的构造函数。您只需要传递变量，而不传递类型。

这是正确编写它的方法:

teacher(int code1,char *name1,char *sub1,char *pub1):
        staff(code1, name1) {

您所有其他错误都是由同一错误引起的，因此您需要对所有派生类的构造函数执行此操作。