Thread: [Big Endian vs Little Endian] Review Problem

Assuming sizeof(int) = 4 ::

My question is what 'y' would be following the code

int x = 0x1234567;
char *p = (char *)&x;
int y = (int)(*p);

(a) On a little endian machine
(b) On a big endian machine

---------------------------------------------------------

The answers are 103 for little endian and
1 for big endian.


However, I am confused how those are the answers. I understand in a little endian machine that the bits are stored from the least significant to the most, and vice-versa for big endian. Should I be converting x into binary at any point? And what is the purpose of the pointer being assigned to the address of x then being casted into an int? Any help would be nice, a little direction for me may help it click in my head.

Originally Posted by Roadrunner2015

Should I be converting x into binary at any point?

Notice that the integer literal is expressed in hexadecimal notation, yet the answers were given in decimal representation. I would expect a conversion from hexadecimal representation to decimal representation instead of from hexadecimal to binary.

Originally Posted by Roadrunner2015

And what is the purpose of the pointer being assigned to the address of x then being casted into an int?

The idea is to examine the bytes of the int by reinterpreting it as an array of char. So, the cast of the address of x to char* reinterprets x as an array of char, where p points to the first char in this array. The cast is a cast of *p, not of p, i.e., to consider the first char in this array as an int.

Convert the value of 0x12 to decimal.
Convert the value of 0x67 to decimal.

And, recheck your answer for big endian because 1 is wrong.

Tim S.

Originally Posted by stahta01

Convert the value of 0x12 to decimal.
Convert the value of 0x67 to decimal.

And, recheck your answer for big endian because 1 is wrong.

0x1234567 can be expressed as 0x01234567 in order for all four bytes in the integer literal to become apparent.

Originally Posted by Roadrunner2015

However, I am confused how those are the answers. I understand in a little endian machine that the bits are stored from the least significant to the most, and vice-versa for big endian.

Endian-ness determines the order of the bytes, not the bits. Since your initial value can be written as 0x01234567, and assuming your int type is big enough to contain that number, the most significant byte is 0x01, and the least is 0x67.

Originally Posted by Roadrunner2015

Should I be converting x into binary at any point?

Nope. Completely unnecessary.

Originally Posted by Roadrunner2015

And what is the purpose of the pointer being assigned to the address of x then being casted into an int?

A char, in C, is generally represented as a single byte. So taking the address of x and storing that in a char pointer, gives us a char pseudo-array of size 4 (the size is an assumption, in this example, for the sake of clarity). Therefore, dereferencing the pointer gives us the first element of the "array." We then convert that value to an int, and store it in y. It will give us the value at the lowest address where x resides. On a little-endian machine, it'll be the least significant byte, and on a big-endian machine, it'll be the most significant byte.

deleted

Originally Posted by Roadrunner2015

The machines, I assume, write AND read in the same direction

A machine will always start reading OR writing, for a particular multi-byte value, at the lowest address of the value in memory. How it's handled after that is architecure-specific.