[SOLVED] Code to find endianness—pointer typecasting

Code to find endianness—pointer typecasting

I was trying to search for code to determine the endianness of the system, and this is what I found:

int main()
{
    unsigned int i = 1;
    char *c = (char *)&i;
    if (*c) {
        printf("Little Endian\n");
    } else {
        printf("Big Endian\n");
    }
}

How does this code work? More specifically, why is the ampersand needed here in the following typecast?

char *c = (char *)&i;

What is getting stored into the pointer c? Does the value i contain it or is the actual address i contained in it? Also, why is this a char for this program?

Solution

While dereferencing a character pointer, only one byte is interpreted (assuming a char variable takes one byte). And in little-endian mode,the least-significant-byte of an integer is stored first.So for a 4-byte integer, say 3, it is stored as

00000011 00000000  00000000  00000000

while for big-endian mode it is stored as:

00000000  00000000  00000000  00000011

So in the first case, the char* interprets the first byte and displays 3, but in the second case it displays 0.

Had you not typecasted it as:

char *c = (char *)&i;

it would have shown a warning about incompatible pointer type. Had c been an integer pointer, dereferencing it would get an integer value 3 irrespective of the endianness, as all four bytes will be interpreted.

NB: You need to initialize the variable i to see the whole picture. Else a garbage value is stored in the variable by default.