Does this code check for endianess ?

Question

Hi

I heard in little endian, the LSB is at starting address and in Big endian MSB is at starting address. SO I wrote my code like this. If not why ?

void checkEndianess()
{

int i = 1;
char c = (char)i;

if(c)
        cout<<"Little Endian"<<endl;
else
    cout<<"Big Endian"<<endl;


}

Answer 1

No, you're taking an int and are casting it to a char, which is a high-level concept (and will internally most likely be done in registers). That has nothing to do with endianness, which is a concept that mostly pertains to memory.

You're probably looking for this:

int i = 1;
char c = *(char *) &i;

if (c) {
   cout << "Little endian" << endl;
} else {
   cout << "Big endian" << endl;
}

Answer 2

Try this instead:

int i = 1;
if (*(char *)&i)
    little endian
else
    big endian

Answer 3

An (arguably, of course ;-P) cleaner way to get distinct interpretations of the same memory is to use a union:

#include <iostream>

int main()
{
    union
    {
        int i;
        char c;
    } x;
    x.i = 1;
    std::cout << (int)x.c << '\n';
}

BTW / there are more variations of endianness than just big and little. :-)