Question

32-bit 2016033010-1339F4A16 Address 00010 11 Big Endian Little Endian 4A 01 9F 01 4A 9F

need help understanding how to convert decimal numbers like these to hexadecimal to out in Big Endian and Little Endian formats

Answer 1

In order to convert decimal number to hexadecimal number, one of the technique is to convert this number to binary and then pair the 4 consecutive binary digit to form a single hexadecimal number.

For example the decimal number 20160330 in binary is 1001100111001111101001010 . Now lets segregate this binary number into chunks of 4 consecutive binary from lest significant field to most significant field.

Hence the binary number with group of 4 bits together will be like below

0001 0011 0011 1001 1111 0100 1010

Here we have added 3 zeros in most significant position to combine 4 bits together.

Now see the decimal number they represent. Any 4 bit binary number will range from 0 to 15.

Hence the binary number 0001 represent 1 , 0011 represent 3,  0011 represent 3, 1001 represent 9, 1111 represent 15 which will be written as hexadecimal digit F, then 0100 represent 4, 1010 represent 10 which in hexadecimal digit is A.

Hence the number in hexadecimal is 13 39 F4 A which is equivalent to 01 33 9F 4A.

Now 1 Byte memory can store 2 digits of hexadecimal.

In Big Endian format, digit from left to right(most significant to least significant) resides in lower to higher address. Hence address 00 contains two digits which is 01, then address 01 contains 33, then address 10 contains 9F and then address 11 contains 4A.

In Little Endian format, digit from left to right(most significant to least significant) resides in higher to lower address. Hence address 00 contains two digits which is 4A, then address 01 contains 9F, then address 10 contains 33 and then address 11 contains 01.

Please comment for any clarification.