Question

I’m new to c++, I have a question about pointer and array

#include <stdlib.h>
#include <stdio.h>

int main(int argc, char * argv[]) 
{
    int a[5] = {1,7,-1,0,2};
    int b[5] = {7,5,3,4,2};
    int c[5] = {1,4,5,3,1};

    *(&a[2] + (&c[5] - 7 - &c[0]))= a[1];
    *(&a[3] + (&c[5] - 7 - &c[0]))= b[1];
    *(&a[4] + (&c[5] - 7 - &c[0]))= c[1];
    int x = a[0] - a[1] + a[2] - a[3];


    printf ("%d", x);
    return 0;
}

Why is that x = 6? thx

Answer 1

Beginning with:

*(&a[2] + (&c[5] - 7 - &c[0]))= a[1];

First we notice the subexpression (&c[5] - 7 - &c[0]) is used multiple times.
(&c[5] - 7 - &c[0]) can be rearranged as (&c[5]- &c[0] - 7 ), which is the address of the fifth int of c, minus the address of the 0th int of c, which results in 5*, so the expression is (5-7) or -2.
(&a[2] -2) is the address of the second index minus two, which is the same as &a[0]. So *(&a[2] + (&c[5] - 7 - &c[0])) is a[0].
By extrapolation, the rest of the code is as follows:

              //a is {1,7,-1,0,2};
a[0] = a[1];  //a is {7,7,-1,0,2};
a[1] = b[1];  //a is {7,5,-1,0,2};
a[2] = c[1];  //a is {7,5, 4,0,2};

So then we get to the final equation:

int x = a[0] - a[1] + a[2] - a[3];

and when we find the values in those positions of the array:

int x = 7 - 5 + 4 - 0;

Which results in 6.

*the addresses are not actually 5 apart (they’re usually 20 or 40), but when you do subtraction of int pointers, it results in the number of ints between them, which is five.