Why can't write a character to a string?

Question

C++ / C#

The 9th line of the code gives the error "Access violation at address".

#include <iostream>
#include <conio.h>
using namespace std;

void main()
{
  char * str1 = new char[6];
  str1 = "Lorem";
  str1[5] = '\0';
  char * str2 = "Lorem\0";

  printf("strlen(str1) = %d\tstrlen(str2) = %d\n", strlen(str1), strlen(str2));
  if (str1 == str2)
  {
    cout << str1 << " == " << str2;
  }
  else
  {
    cout << str1 << " != " << str2;
  }
  _getch();

  delete[] str1, str2;
}

I can't understand these things:
1) why can't I write a character (at any index)?
2)why does the expression "str1 == str2" return false if the strings are identical?
3) why in this code the character "\0" is normally written in all lines, but in the line "data[6][faculty_s] = '\0';" the same error?

#include <iostream>
#include <conio.h>
#define MAX_LOGIN_SIZE 50
#define MAX_PASSWORD_SIZE 50
#define MAX_SURNAME_SIZE 32
#define MAX_NAME_SIZE 32
#define MAX_PATRONYMIC_SIZE 32
#define GROUP_NAME_SIZE 8//example "ECE-1511"
#define MAX_FACULTY_SIZE 21//"System administration"
using namespace std;

void main()
{
  FILE * accounts_w = nullptr;
  fopen_s(&accounts_w, "Accounts.dat", "wb");
  
  char * str = "Anto_cn69";
  short size = strlen(str);
  fwrite(&size, sizeof(size), 1, accounts_w);
  fputs(str, accounts_w);

  str = "4utc1";
  size = strlen(str);
  fwrite(&size, sizeof(size), 1, accounts_w);
  fputs(str, accounts_w);

  str = "Antonyuk";
  size = strlen(str);
  fwrite(&size, sizeof(size), 1, accounts_w);
  fputs(str, accounts_w);

  str = "Arkadiy";
  size = strlen(str);
  fwrite(&size, sizeof(size), 1, accounts_w);
  fputs(str, accounts_w);

  str = "Olegovich";
  size = strlen(str);
  fwrite(&size, sizeof(size), 1, accounts_w);
  fputs(str, accounts_w);

  int age = 18;
  fwrite(&age, sizeof(age), 1, accounts_w);

  str = "ECE-1511";
  size = strlen(str);
  fwrite(&size, sizeof(size), 1, accounts_w);
  fputs(str, accounts_w);

  str = "Programming";
  size = strlen(str);
  fwrite(&size, sizeof(size), 1, accounts_w);
  fputs(str, accounts_w);

  _getch();

  fclose(accounts_w);
  delete[] str;

  FILE * accounts = nullptr;//содержит данные учётных записей студентов
  fopen_s(&accounts, "Accounts.dat", "rb");

  char ** data = new char*[7];
  data[0] = new char[MAX_PASSWORD_SIZE + 1];
  data[1] = new char[MAX_SURNAME_SIZE + 1];
  data[2] = new char[MAX_NAME_SIZE + 1];
  data[3] = new char[MAX_PATRONYMIC_SIZE + 1];
  data[4] = new char[3];//2 цифры + '\0'
  data[5] = new char[GROUP_NAME_SIZE + 1];
  data[6] = new char[MAX_FACULTY_SIZE + 1];
  char temp_login[MAX_LOGIN_SIZE + 1];
  short log_s, pass_s, surn_s, name_s, patr_s, faculty_s;//_s - "size"

  fread(&log_s, sizeof(log_s), 1, accounts);
  fread(temp_login, sizeof(char), log_s, accounts);

  //пароль
  fread(&pass_s, sizeof(pass_s), 1, accounts);
  fread(data[0], sizeof(char), pass_s, accounts);
  data[0][pass_s] = '\0';

  //фамилия
  fread(&surn_s, sizeof(surn_s), 1, accounts);
  fread(data[1], sizeof(char), surn_s, accounts);
  data[1][surn_s] = '\0';

  //имя
  fread(&name_s, sizeof(name_s), 1, accounts);
  fread(data[2], sizeof(char), name_s, accounts);
  data[2][name_s] = '\0';

  //отчество
  fread(&patr_s, sizeof(patr_s), 1, accounts);
  fread(data[3], sizeof(char), patr_s, accounts);
  data[3][patr_s] = '\0';

  //возраст
  fread(&age, sizeof(age), 1, accounts);
  data[4][2];

  //группа
  fread(data[5], sizeof(char), GROUP_NAME_SIZE, accounts);
  data[5][GROUP_NAME_SIZE] = '\0';

  //факультет
  fread(&faculty_s, sizeof(faculty_s), 1, accounts);
  fread(data[6], sizeof(char), faculty_s, accounts);
  data[6][faculty_s] = '\0';

  for (short i = 0; i < 7; i++)
  {
    if (i == 4 || i == 6)
    {
      continue;
    }
    cout << data[i] << endl;
  }
  _getch();

And one more question: how can you still convert int to char* (because the system swears at "_itoa_s(age,data[4], 2, 10);", and the compiler does not skip previous versions of this function)?

Answer 1

Go .
Char * str =blbablabla
Here, firstly, and so at the end there will be zero.
If not, then in this line you can immediately add zero. Vrchd whether this will be necessary.
The memory area is read-only.
2. Google comparison of strings and pointers
3.t am allocated memory for reading and writing
And then apparently they went abroad. hard to see on mobile.
4. You should only use safe options and read other things about options. I won't tell now

Answer 2

char * str1 = new char[6]; // тут вы выделяете 6 байт памяти
                                         // и присваиваете адрес выделенной памяти переменной str1
str1 = "Lorem"; // тут вы теряет ранее полученный адрес (утечка памяти)
                        // и переменной str1 присваиваете новый адрес из (not write access) блока памяти.
str1[5] = '\0';   // тут у вас уже проблема, так как на эту область памяти стоит защита от записи.

The contents of the string must be copied to the allocated memory