Answer :
The content of the DL register after executing the given assembly language code will be 17H. If your system isn't little-endian, the answer might vary, but this should give you a solid understanding of the steps involved.
Firstly, let's understand the initial setup. The DATA defined by DW (Define Word) is an array that contains four 16-bit values: 567FH, 7BC0H, 429CH, and 0017H. Each of these occupy 2 bytes (since they are 16-bit). The offset address of DATA is given as 100H.
Next, the instruction MOV BP, OFFSET DATA is storing the offset address of DATA in the register BP. So now, BP contains 100H.
MOV SI, 3 is simply storing the value 3 into SI register.
Finally, the MOV DL,[BP][SI] instruction. This instruction is saying to move the contents at the address calculated by (BP + SI) into DL. In our case, BP is 100H (the offset of DATA), and SI is 3.
Remember that assembly language uses byte addressing. In our DATA array, each element takes up 2 bytes (because we're dealing with 16-bit words). Hence, moving 3 locations from the base address 100H means moving 3*2 = 6 bytes away from the base. The effective address is therefore 100H + 6 = 106H.
So, if we look at our DATA array considering each element occupies 2 bytes:
- At 100H and 101H, we have 567FH
- At 102H and 103H, we have 7BC0H
- At 104H and 105H, we have 429CH
- At 106H and 107H, we have 0017H
You'll see that the 106H address contains the low byte of the 0017H value from our DATA array. If we assume a little-endian memory system (which is common, and means that the least significant byte is at the lower address), the content of DL will be 17H, as this is the low byte of the word at that address.
To summarize, the content of the DL register after executing the given assembly language code will be 17H. If your system isn't little-endian, the answer might vary, but this should give you a solid understanding of the steps involved.
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