Assume that memory cells 60 and 61 and register R currently...

Assume that memory cells 60 and 61 and register $R$ currently have the following values: Register R: 13 60: 472 61: $-1$ Using the instruction set in Figure $6.5$, what is in register $R$ and memory cells 60 and 61 after completion of each of the following operations? Assume that each instruction starts from the above conditions. a. LOAD 60 b. STORE 60 c. ADD 60 d. COMPARE 61 e. IN 61 (Assume that the user enters 50.) f. OUT 61

Assume that memory cells 60 and 61 and register $R$ currently have the following values:
Register R: 13
60: 472
61: $-1$
Using the instruction set in Figure $6.5$, what is in register $R$ and memory cells 60 and 61 after completion of each of the following operations? Assume that each instruction starts from the above conditions.
a. LOAD 60
b. STORE 60
c. ADD 60
d. COMPARE 61
e. IN 61 (Assume that the user enters 50.)
f. OUT 61

Key Concepts

Comparison Instruction

The compare instruction is used to evaluate the relationship between two numerical values, often by subtracting one value from another without changing the actual data. The result typically influences program flow, such as determining branch decisions in conditional statements.

Registers

Registers are small, high-speed storage locations within the CPU that temporarily hold data and instructions during program execution. They are crucial for performing operations quickly and are typically used to store operands and the results of computations.

Memory

Memory refers to the system’s storage where data and instructions are kept for processing. It is organized in addressable units (cells) that can hold individual pieces of data, and it plays a key role in how the CPU retrieves and stores information.

Load Instruction

The load instruction is used to transfer data from a specified memory address into a CPU register. This operation is fundamental in assembly language programming as it allows the CPU to work on data that was previously stored in memory.

Store Instruction

The store instruction moves data from a CPU register to a designated memory location. This operation is essential for saving results of computations or updating memory with new data generated during program execution.

Arithmetic Operations in Assembly

Arithmetic operations, such as addition, are performed by modifying the contents of registers. These instructions take one or more operands, perform the specified mathematical operation, and place the result back into a register, making them vital for numerical computations.

Input/Output Instructions

Input/Output (I/O) instructions manage the interaction between the computer and external devices. Input instructions bring data from devices (like keyboards) into the system, while output instructions send data from the CPU to devices (like displays or printers), facilitating user interaction and external communications.

Transcript

0:00 Hello.

00:02 So in this question we're asked to translate some instructions from english into machine language code in the machine described in the appendix of this book.

00:14 So in a, we need a code to load register number 6 with the hexadicimal value of 77.

00:27 So that's the first task here.

00:32 So we need to load a register with a bit string.

00:36 So this is the op code number two, and we need to load the register 6, and then the value that we need to load the register with is entered here.

00:49 So 2677 is the instruction code that will load register 6 with the value 77.

00:56 Now let's move to b.

00:58 Our goal here is to again load but now register 7 with the contents of memory cell with the address of 77.

01:13 So this is different from the previous one because in that case we would use the upcode 1 because this is the one used to load a register with the content of a certain memory cell.

01:25 So here load register 7 with the bit string located at 7 .7.

01:34 I don't know what will be stored now in register 7 because this depends on the content of the memory cell number 77 itself.

01:42 So that's the code in that case.

01:45 Now, c, we need to jump to the instructions located at the memory cells.

01:56 The memory cells with 24 if the content of register 0 is equal to the content in register a.

02:11 So now we need the jump up code and we know that the jump of code is b.

02:18 This is the conditional jump code and it compares whatever register here with the register number 0.

02:28 So now we need to compare a, the register a, with the zero...