There is the following f-I.
int main() { bool arr[6]; for (int i = 0; i < 6; i++) { cout << &arr[i] << " "; } cout << endl; char arr2[6]; for (int i = 0; i < 6; i++) { cout << &arr2[i] << " "; } cout << endl; return 0; } Why, when an array of char displays incomprehensible characters, and not hexadecimal numbers?
cout << c ; where с is of type char* , it is assumed that с is a null-terminated sishnaya string. To output an address of type char* , you need to cast this type to void* , for example: cout << (void*)c;
(void*) and in most cases everything will be as you expect. Perhaps (this needs to be checked) there will be problems with the addresses of the functions of the classes (some call this "methods", but this is not correct - there are no methods in c ++). - KoVadimI add to the above - this is how you can get the addresses of the cells of the array of type char:
#include <iostream> using namespace std; int main(){ bool arr[6]; for (int i = 0; i < 6; i++) { cout << &arr[i] << " "; } cout << endl; char arr2[6]; for (int i = 0; i < 6; i++) { cout << &arr2 + i << " "; // получаем адреса с помощью адресной арифметики } cout << endl; return 0; } 
Source: https://ru.stackoverflow.com/questions/391056/
All Articles