Solved In Executing The Following Code Start Mov Si 0300 Chegg Here’s the best way to solve it. the result will be: " (0300h) = 1ah, (0301h) = 2bh, (0302h) = 3ch, (0303h) = 4dh" explanation: "the code above will move the value at memory location 0300h (a9h) into the si register. it will then move the value at memory location 0302h (5bh) into …. in executing the following code: start: mov si, 0300 mov. Answer to solved in executing the following code: start: mov si, 0300 | chegg . this problem has been solved! in executing the following code: start: mov.
Solved In Executing The Following Code Start Mov Si 0300 Chegg The result of executing the given code will be: (0300h)=1ah, (0301h)=2bh, (0302h)=56h, (0303h)=78h. what are the intermediate steps involved in executing the code? to understand the result, let's break down the execution of the code step by step: 1. mov si, 0300: this instruction moves the value 0300h into the si register. 2. The code above loads 0002 into cx, 0300 into si, and 0302 into di. in the next step, mov al, (si) xor (di), al will take the contents of the si memory location, perform an exclusive or operation with the di memory location contents, store the results in the al register. then the si, di, and cx registers are updated by 1, and 2 is subtracted. The x86 64 assembly mov instruction operates within specific parameters and boundaries. it’s important to be aware of these, as they guide the effective usage of this command. to start, the source and destination data sizes must match. whether you’re dealing with 8, 16, 32, or 64 bits, the sizes must be the same. The brackets mean "memory access". your code would have the processor fetch a value from memory ( [bx]) and store it into memory ( [si] ), but this is not a valid encoding for the mov instruction. in fact, there is no single x86 instruction that performs such a memory memory operation. you have to split it into two parts, like so: mov ax, [bx.
Solved When Executing The Following Code Start Mov Si Ooh The x86 64 assembly mov instruction operates within specific parameters and boundaries. it’s important to be aware of these, as they guide the effective usage of this command. to start, the source and destination data sizes must match. whether you’re dealing with 8, 16, 32, or 64 bits, the sizes must be the same. The brackets mean "memory access". your code would have the processor fetch a value from memory ( [bx]) and store it into memory ( [si] ), but this is not a valid encoding for the mov instruction. in fact, there is no single x86 instruction that performs such a memory memory operation. you have to split it into two parts, like so: mov ax, [bx. Add bx, 0008h the register ax contains some value which needs to be stored at a location as follows: mov [bx], ax calculate the address at which the value of the ax will be stored. answer. after executing the first instruction, the value of bx register is as follows: bx = 3040h. the bx register is an offset of the data segment (ds) register. When executing the following code: start: mov si, ooh mov cx, 5 answered step by step. solved by verified expert.
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