converting char array to int using c++

Question

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;;
}

I read that? Why? The first one trying to use this in my code will have to defend herself/himself. — JVApen
– JVApen, Commented May 18, 2019 at 13:42

Artyer · Accepted Answer · 2019-05-18 11:50:12Z

2

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.

answered May 18, 2019 at 11:50

Artyer

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