Question

I’m confused when I was reading an article about Big/Little Endian.

Code goes below:

#include <iostream>
using namespace std;

int i = 12345678;

int main()
{
    char *p = (char*)&i;  //line-1

    if(*p == 78)  //line-2
        cout << "little endian" << endl;
    if(*p == 12)
        cout << "big endian" << endl;

}

Question:

1. In line-1, can I do the conversion using static_cast<char*>(&i)?

2. In line-2, according to the code, if it’s little-endian, then 78 is stored in the lowest byte, else 12 is stored in the lowest byte. But what I think is that, i = 12345678; will be stored in memory in binary.

   If it’s little-endian, then the last byte of i‘s binary will be stored in the lowest byte, but what I don’t understand is how can it guarantee that the last byte of i is 78?

   Just like, if i = 123;, then i‘s binary is 01111011, can it guarantee that in little-endian, 23 is stored in the lowest byte?

Answer 1

1. I’d prefer a reinterpret_cast.

2. Little-endian and big-endian refer to the way bytes, i.e. 8-bit quantities, are stored in memory, not two-decimal quantities. If i had the value 0x12345678, then you could check for 0x78 and 0x12 to determine endianness, since two hex digits correspond to a single byte (on all the hardware that I’ve programmed for).