In the following memory dump file there is a 6 word array st
Solution
Given image contains the memory dump from the address of CS:270 and here each byte holds two digits (4 bits +4 bits). Now considering the value at address CS:0286 = 14 and from here we need to consider 6 Words and as we already know that 1 Word = 2 Bytes and so 6 Words includes all the location values from
CS:0286, CS:0287, CS:0288, CS:0289, CS:028A, CS:028B, CS:028C, CS:028D, CS:028E, CS:028F, CS:0290, CS:0291
=> 14 00 1D 00 F2 FF 23 00 FC FF 1B 00
These values which are considered as 6 words (6 *2 = 12 Bytes) from the address CS:0280 and now these values are to be converted to decimal from the hexadecimal notation..
Converting hexadecimal number to decimal:
consider XY is the number represented in hexadecimal then to convert it into decimal first write the 4-bit representation for each digit and then convert the obtained binary value into decimal by multiplying with powers of 2
CS:0286 value= 14 in hexadecimal notation
0001 0100 in binary notation
2^7*0+2^6*0+2^5 *0+2^4 *1 +2^3 *0+ 2^2 *1+2^1 *0+2^0 *0
=16+4=20 in decimal notation
CS:0287 value=00 in hexadecimal notation
0000 0000 in binary
00 in decimal notation
CS:0288 value=1D in hexadecimal notation
0001 1101 in binary notation
2^7*0+2^6*0+2^5 *0+2^4 *1 +2^3 *1+ 2^2 *1+2^1 *0+2^0 *1
=29
CS:0289 value =00 in hexadecimal notation
0000 0000 in binary notation
00 in decimal notation
Cs:028A value=F2 in hexadecimal notation
11110010 in binary notation
2^7*1+2^6*1+2^5 *1+2^4 *1 +2^3 *0+ 2^2 *0+2^1 *1+2^0 *0
242 in decimal notation
CS:028B value=FF in hexadecimal notation
1111 1111 in binary notation
2^7 *1+2^60 *1+2^5 *1+2^4 *1 +2^3 *1+ 2^2 *1+2^1 *0+2^0 *1
255 in decimal notation
CS:028C value=23 in hexadecimal notation
0010 0011 in binary notation
2^7*0+2^6*0+2^5 *1+2^4 *0 +2^3 *0+ 2^2 *0+2^1 *1+2^0 *1
35 in decimal notation
CS:028D value=00 in hexadecimal
0000 0000 in binary
00 in decimal
CS:028E value=FC in hexadecimal
1111 1100 in binary
2^7*1+2^6*1+2^5 *1+2^4 *1 +2^3 *1+ 2^2 *1+2^1 *0+2^0 *0
252 in decimal
CS:028F value=FF in hexadecimal
1111 1111 in binary
2^7*1+2^6*1+2^5 *1+2^4 *1 +2^3 *1+ 2^2 *1+2^1 *1+2^0 *1
255 in decimal
CS:0290 value= 1B in hexadecimal
0001 1011 in binary
2^7*0+2^6*0+2^5 *0+2^4 *1 +2^3 *1+ 2^2 *0+2^1 *1+2^0 *1
27 in decimal
Cs:0291 value=00 in hexadecimal
0000 0000 in binary
00 in decimal
So all the 6 words from address location CS:0286 are converted into decimal values

