FILE1

FILE 2

Question:

data offset: Ox0=0 00 01 02 03 04 05 06 07 08 09 Oa Ob Oc Od de of 0000000-0 38 00 00 00 6a 6f 68 6e 20 73 60 69 74 68 00 00 06 00 00 00 2e 00 00 00 60 61 72 79 20 6a 6f 6e 0000000-do 65 73 00 00 0c 00 00 00 24 00 00 00 74 69 6d 20 0000000-00 0000000-be 64 61 76 69 73 00 00 00 05 00 00 00 la 00 00 00 0000000-an 70 61 6d 20 63 60 61 72 6b 00 00 00 0a 00 00 00 hexg Ox38 8... john smith.. mary jon es. .$... tim davis.. pam clark. ints 56 int16 56 int32 56 int64 56 uints 56 uint16 56 uint32 56 uint64 56 float double big endian data offset: Ox0=0 .... 00 01 02 03 04 05 06 07 08 09 Oa Ob Oc Od Oe Of 0000000-eo 00 00 00 38 6a 6f 68 6e 20 73 6d 69 74 68 00 00 00 00 00 06 00 00 00 2e 60 61 72 79 20 6a 6f 6e 0000000-do 65 73 00 00 00 00 00 OC 00 00 00 24 74 69 6d 20 0000000-CO 64 61 76 69 73 00 00 00 00 00 00 05 00 00 00 la 0000000-be 0000000-a 70 61 6d 20 63 60 61 72 6b 00 00 00 00 00 00 ea hexs Ox0 ...Bjohn smith.. mary jon $tim davis... pam clark. es. ints 0 int16 0 int32 0 int64 0 uint 0 INICI uint16 0 uint32 0 uint64 0 float double big endian structure Wie search checksum question. Open these binary files using your favorite hexadecimal editor. Both these files are 80 bytes in length and contain employee records as shown in the data structure below: struct employee { data_typel age; 4 bytes (int) data_type2 name [12]; character array of size 12 data_type3 division; 4 bytes }; Using the knowledge of common data type sizes in the C programming language (as shown above), I'd like you to guess the actual data structure in the given files and the number of employee records in each file. Also, identify which file is little-endian and which is big-endian. Justify your answers. Explain the need for these different "endian" schemes. How do two machines with different endian-ness attributes communicate over a network? Lastly, while the files are in reality 80 bytes in size, as you can verify, how is it that the size on disk is much larger (about 4KB = 4096 bytes on some machines) and 0 bytes on Windows 10? data offset: Ox0=0 00 01 02 03 04 05 06 07 08 09 Oa Ob Oc Od de of 0000000-0 38 00 00 00 6a 6f 68 6e 20 73 60 69 74 68 00 00 06 00 00 00 2e 00 00 00 60 61 72 79 20 6a 6f 6e 0000000-do 65 73 00 00 0c 00 00 00 24 00 00 00 74 69 6d 20 0000000-00 0000000-be 64 61 76 69 73 00 00 00 05 00 00 00 la 00 00 00 0000000-an 70 61 6d 20 63 60 61 72 6b 00 00 00 0a 00 00 00 hexg Ox38 8... john smith.. mary jon es. .$... tim davis.. pam clark. ints 56 int16 56 int32 56 int64 56 uints 56 uint16 56 uint32 56 uint64 56 float double big endian data offset: Ox0=0 .... 00 01 02 03 04 05 06 07 08 09 Oa Ob Oc Od Oe Of 0000000-eo 00 00 00 38 6a 6f 68 6e 20 73 6d 69 74 68 00 00 00 00 00 06 00 00 00 2e 60 61 72 79 20 6a 6f 6e 0000000-do 65 73 00 00 00 00 00 OC 00 00 00 24 74 69 6d 20 0000000-CO 64 61 76 69 73 00 00 00 00 00 00 05 00 00 00 la 0000000-be 0000000-a 70 61 6d 20 63 60 61 72 6b 00 00 00 00 00 00 ea hexs Ox0 ...Bjohn smith.. mary jon $tim davis... pam clark. es. ints 0 int16 0 int32 0 int64 0 uint 0 INICI uint16 0 uint32 0 uint64 0 float double big endian structure Wie search checksum question. Open these binary files using your favorite hexadecimal editor. Both these files are 80 bytes in length and contain employee records as shown in the data structure below: struct employee { data_typel age; 4 bytes (int) data_type2 name [12]; character array of size 12 data_type3 division; 4 bytes }; Using the knowledge of common data type sizes in the C programming language (as shown above), I'd like you to guess the actual data structure in the given files and the number of employee records in each file. Also, identify which file is little-endian and which is big-endian. Justify your answers. Explain the need for these different "endian" schemes. How do two machines with different endian-ness attributes communicate over a network? Lastly, while the files are in reality 80 bytes in size, as you can verify, how is it that the size on disk is much larger (about 4KB = 4096 bytes on some machines) and 0 bytes on Windows 10