Verifying Endianness

[3246251196]

I believe are linux commands that can output the value of endianness.

Would it also be true to assume that a bitwise shift left will indicate the value of endianness? Where given a integral value value, v, and a number of left shifts, n, if (v * n^2) > v this implies Big Endian and vice versa?

[zobayer1]

I don't know about commands, but this code should determine the endianness

#define BIG_ENDIAN 0
#define LITTLE_ENDIAN 1

int TestByteOrder() {
    short int word = 0x0001;
    char *byte = (char *) &word;
    return (byte[0] ? LITTLE_ENDIAN : BIG_ENDIAN);
}

[Bachstelze]

Would it also be true to assume that a bitwise shift left will indicate the value of endianness? Where given a integral value value, v, and a number of left shifts, n, if (v * n^2) > v this implies Big Endian and vice versa?

If you mean in C, it's a big no! All operators in C work on values, not bits. (1 << 5) will always be 32, regardless of endianness. One way to test for it is:

unsigned int n = 1;
bool islittle = *((unsigned char*)(&n)) == 1;

[3246251196]

Thank you for your two replies. Identical.

We are saying:
1- Here is a 4 byte integer
2- Let the first byte (lowest address) of the memory that stores integer be converted to a character (so as to represent 1 byte)
3- If this byte, when de-referenced, contains the value 1, then we are working with little, else big.

[zobayer1]

Thank you for your two replies. Identical.

We are saying:
1- Here is a 4 byte integer
2- Let the first byte (lowest address) of the memory that stores integer be converted to a character (so as to represent 1 byte)
3- If this byte, when de-referenced, contains the value 1, then we are working with little, else big.

Yes, and Bachstelze explained why bit shifting won't work, even if they seem correct theoretically.