Can someone explain this "endian-ness" function for me?

Question

Write a program to determine whether a computer is big-endian or little-endian.

bool endianness() {
     int i = 1;
     char *ptr;
     ptr  = (char*) &i;
     return (*ptr);
}

So I have the above function. I don't really get it. ptr = (char*) &i, which I think means a pointer to a character at address of where i is sitting, so if an int is 4 bytes, say ABCD, are we talking about A or D when you call char* on that? and why?

Would some one please explain this in more detail? Thanks.

So specifically, ptr = (char*) &i; when you cast it to char*, what part of &i do I get?

Answer 1

If you have a little-endian architecture, i will look like this in memory (in hex):

01 00 00 00
^

If you have a big-endian architecture, i will look like this in memory (in hex):

00 00 00 01
^

The cast to char* gives you a pointer to the first byte of the int (to which I have pointed with a ^), so the value pointed to by the char* will be 01 if you are on a little-endian architecture and 00 if you are on a big-endian architecture.

When you return that value, 0 is converted to false and 1 is converted to true. So, if you have a little-endian architecture, this function will return true and if you have a big-endian architecture, it will return false.

Answer 2

If ptr points to byte A or D depends on the endianness of the machine. ptr points to that byte of the integer that is at the lowest address (the other bytes would be at ptr+1,...).

On a big-endian machine the most significant byte of the integer (which is 0x00) will be stored at this lowest address, so the function will return zero.

On a litte-endian machine it is the opposite, the least significant byte of the integer (0x01) will be stored at the lowest address, so the function will return one in this case.