CODEWITHSUNDEEP
c++arrayscharint
Hadi
Hadi

Reputation: 574

converting char array to int using c++

Can anyone explain me how this code results 16843009? How it works?

As I saw in my tests, (int *)&x results 0x61ff1b and as I know that is the address of the first element in the array. and how the result of *(int *)&x is 16843009? Thanks. 

#include <iostream>

using namespace std;


int main()
{
    char x[5] = {1, 1, 1, 1, 1};
    cout << *(int *)&x;

   return 0;;
}

Upvotes: 0

Views: 126

Answers (1)

Artyer
Artyer

Reputation: 40951

If we write 16843009 as binary we get 1000000010000000100000001. Padding that with extra zeros we get: 00000001000000010000000100000001. Every 8 bits (which is a char) has a value of 00000001, which is 1.

&x is a pointer to an array of char (Specifically a char(*)[5]). This is reinterpreted as a pointer to int. On your system, int is probably 4 bytes, and all four of those bytes are seperately set to 1, which means you get an int where every 8 bits are set to 1.

Upvotes: 2

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