Ｃ++ 配列のポインタ管理について

Question

###前提・実現したいこと
配列で定義されたポインタで、各要素ごとの配列を管理したいと考えております。
下記のようにすることは不可能でしょうか？

someone[i] = w[i];
cout << someone[i].salary << "\n";

###発生している問題・エラーメッセージ
Main.cpp:7:29: warning: ISO C++11 does not allow conversion from string literal to 'char ' [-Wwritable-strings]
char Worker::CompanyName = "ABC";
^
Main.cpp:10:8: warning: expression result unused [-Wunused-value]
this->num;

Main.cpp:43:18: error: assigning to 'Worker *' from incompatible type 'Worker'; take the address with &
     someone[i] = w[i];
                ^ ~~~~
                  &
Main.cpp:44:25: error: member reference type 'Worker *' is a pointer; maybe you meant to use '->'?
     cout << someone[i].salary << "\n"; 
             ~~~~~~~~~~^
                       ->
###ソースコード
```＜メイン文＞
#include <iostream>
#include "string.h"
#include "Worker.h"
using namespace std;

int Worker::no = 0;
char* Worker::CompanyName = "ABC";

Worker::Worker(){
this->num;
cout << "-------------------" << Worker::no << "\n";
this->salary = 0;
strcpy(this->department, "NotDefinedDepartment");
strcpy(this->name, "NotDefinedName");
no++;
}

Worker::~Worker(){
}

Worker::Worker(double salary ,char department[100], char name[100]){
 this->num = no;
 this->salary = salary;
 strcpy(this->department,department);
 strcpy(this->name,name);
 no++;
}

void Worker::ShowData(){
   cout << "no = " << this->no << "\n";
   cout << "salary = " << this->salary << "\n";
   cout << "department = " << this->department << "\n";
   cout << "name = " << this->name << "\n";
}

int main(void){
 Worker w[3];
 Worker* someone[3];
 for(int i=0 ; i<3 ;i++){
     w[i].salary = 100;
     strcpy(w[i].department, "E");
     strcpy(w[i].name, "Takayuki");
     someone[i] = w[i];
     cout << someone[i].salary << "\n"; 
     
 }
 
   w[0].ShowData();
   w[1].ShowData();
}

＜ヘッダー文＞
class Worker{
public:
   static int no;
   int num;
   static char* CompanyName;
   double salary;
   char department[100];
   char name[100];
  Worker();
  ~Worker();
 Worker(double salary ,char department[100], char name[100]);
  void ShowData();
  
};
```

###補足情報(言語/FW/ツール等のバージョンなど)
paiza.io

Answer 1

someone[i]がw[i]を示すようにしたいということならw[i]のアドレスを代入する必要があります。

someone[i] = &w[i]; 

また、someoneはポインタなのでアクセスするにはアロー演算子->を使用します。

cout << someone[i]->salary << "\n"; 

Comments

[saito.kaz]

ありがとうございます。