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31.3.4. Displacement Addressing Mode Example
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Today, we’re going to learn about displacement addressing mode. Can anyone explain what displacement means in this context?
Does it mean we get a value from a register and add something to it?
Exactly! In displacement addressing, we calculate an effective address by adding a constant or 'displacement' value to the content of a register. For example, if R1 holds 1 and our displacement is also 1, the effective address is 2. This way, we can access different memory locations.
So, R1 + displacement gives us where in memory we can find our data?
Correct, Student_2! It's essentially how we dynamically calculate which memory location to access while executing instructions.
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Now, let's connect this to registers, specifically R1 and R5 in our example. Can someone remind me what values R1 and R5 start with?
R1 starts with 1, and R5 starts with 0.
Good memory! As we use these registers, R1 continues to increment while R5 accumulates the values from memory locations accessed through R1. If R1 increments by 1 each time, what happens to R5?
R5 will keep adding the values of the items in the array that R1 points to, right?
Exactly! This process shows how effectively we can traverse an array using displacement addressing mode.
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Create a free accountLet's think about the instructions in terms of opcode and operands. What is the opcode responsible for in an instruction?
The opcode tells the CPU what operation to perform, like adding or subtracting.
Exactly! And what about the operand?
The operand gives the values or addresses used with that operation.
Right on target! In our example, how might the opcode and operands interact when we want to do addition using displacement addressing?
The opcode would specify the addition, and the operand would tell where to find the values in memory using the effective address we calculated.
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Create a free accountLet's break down a practical example together. Starting with R1 at 1 and R5 at 0, what is the first thing we do?
We calculate the effective address by adding the contents of R1 and the displacement.
Correct! That gives us the first memory location to access. If the memory contains some data, what should we do next?
We add that data to R5, right?
Yes! And then increment R1. This reiterative loop allows us to accumulate data until all relevant memory locations have been accessed.
So it’s like a loop structure where R1 is the index, and R5 is accumulating a total?
Exactly! Great observation.
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Create a free accountTo recap, displacement addressing allows us to dynamically calculate effective addresses using registers and displacement values. Can anyone summarize how this interaction works?
We start with R1 and add a displacement to get the effective address, which we use to find data in memory, and then update R5!
And as we increment R1, we access different data points in the array!
Perfectly summarized! Understanding this mode is critical to efficient programming and CPU function. Keep exploring!
Overview
Short Summary
This section explains the displacement addressing mode in computer architecture, utilizing examples to illustrate how effective addresses are calculated and used in instructions.
Medium Summary
In this section, we explore the displacement addressing mode by detailing how registers, memory locations, and arithmetic operations interact to execute instructions. Through an example involving registers R1 and R5, we demonstrate the calculation of effective addresses and the process of accumulating values from memory locations.
Detailed Summary
Displacement Addressing Mode Example
This section delves deeply into the concept of displacement addressing mode commonly used in computer architecture, illustrating how effective addresses are computed and utilized within instruction execution. The example given focuses on the operation involving two registers: R1 and R5.
The initial state is explained with R1 set to 1 and R5 set to 0, establishing a clear starting point. The example illustrates how effective addresses are derived by taking the content of R1 and adding a displacement value (in this case, 1). Thus, the effective address is calculated as the content of R1 plus 1, allowing the program to access the first element of an array starting at memory location 1.
As operations are performed, R5 accumulates values from these memory locations, demonstrating a fundamental loop structure akin to summing elements in an array. The subsequent example addresses the implications of instruction size and structure, showing how larger operands might necessitate multi-word instructions and altering the program counter behavior based on instruction lengths.
The section concludes with discussions of similar addressing modes, reinforcing the importance of understanding how displacement addressing is leveraged in programming and hardware design.
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Create a free accountSo, what is this I am effectively trying to do? So, in this case register 𝑅5 will be 𝑅5 plus what is the content of the memory location how can you find out if whatever is 𝑅1 will be added to the content of the memory location how the memory location is calculated it is content of 𝑅1 + 1.
Detailed Explanation
Displacement addressing mode is a method of accessing memory locations using an index register. In this example, the content of register R5 is updated by adding to it the content found at a memory location calculated as R1 + 1. Here, R1 is used as an index, and its value determines which memory location to access. The addition of 1 means that the effective address points to the next location in memory based on the index in R1.
Examples & Analogies
Think of R1 as an index for a library shelf where book placements start from 1. If R1 is 1, it points to the first shelf, and if the content of that shelf is updated, R5 will contain that new value. If you increment R1 to add 1, it points to the second shelf for the next operation.
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Create a free accountSo, initially they are assuming that 𝑅1 is having the value of 1 and 𝑅5 has the value of 0 that is 𝑅5 is reset.
Detailed Explanation
The registers begin with specific initial values: R1 starts at 1, and R5 starts at 0 (reset state). This setup is important as it establishes what values will be used in calculations. When R1 is 1, it points to the first memory location as we previously discussed, and R5 being zero means that anything added to it will solely depend on the value accessed through the displacement at that memory location.
Examples & Analogies
Imagine you have a notebook (register R5) that starts empty (0). Your friend (register R1) tells you to write down the value from the first shelf (1). At first, your notebook has no entries, so the first entry becomes that value.
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Create a free accountSo, what will happen the instruction takes the address 1 and adds to the content of 𝑅1. So, the content of 𝑅1 and 𝑅2 both have 0. So initially both of them has 0 value so that is this 1 will be added to the content of 𝑅1.
Detailed Explanation
The process described continues by taking the address determined by R1, which is currently 1. Both R1 and R2 are initially set to 0, so when we add 1 to R1, it determines that we are retrieving the content from the first memory location, which feeds into the next operation for R5.
Examples & Analogies
Continuing with our library analogy, if the index (R1) is set to point at shelf 1 and it 'retrieves' the first book's details to note down in your empty notebook (R5), you'll now have valuable content filled in.
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Create a free accountThen next that is actually what I have told you whatever is present over here will be added to the contents of 𝑅5 which is now 0... So, 1 + 1 will be 2.
Detailed Explanation
After retrieving and adding the content from memory to R5, the next step involves incrementing R1 by 1. This effectively prepares for the next operation, allowing you to access the second memory location. This process continues in increments, allowing the CPU to read through all memory locations sequentially.
Examples & Analogies
It’s like writing down notes from each book you get from the library shelf. Once you finish one book, you raise the index number to pick the next one on the shelf. If you started with book 1, now you will go for book 2 for the next entry.
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Create a free accountSo, there is something happening like 𝑠 = 𝑠 + 𝑎[𝑖] where 𝑎 is the array and 𝑖 is your instruction.
Detailed Explanation
The example uses an array access pattern where s represents R5 and a[i] refers to the value at the ith position of the array. This highlights how displacement addressing helps manage data within arrays by dynamically updating the base address using the index in R1, making it efficient for operations involving series data.
Examples & Analogies
Imagine you have a sequence of boxes (array) where each box holds a value. You tell your friend (processor) to open box number 1 (R1's value) to see what it has and to add that to your total sum (R5). The sum gets updated to include this new value.
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Key Concepts
Core takeaways and short definitions to help you quickly recall the key ideas from this section.
Displacement Addressing: A technique for calculating memory addresses dynamically.
Effective Address Calculation: Sum of register content and displacement value.
Accumulation in Registers: Using registers like R5 to aggregate values from multiple memory accesses.
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Glossary
Displacement Addressing
A method of addressing where the effective address is calculated by adding a constant value to the content of a register.
Effective Address
The actual memory address calculated from the operands used in the instruction.
Register
A small amount of storage available directly in the CPU used to hold temporary data and addresses.
Opcode
The part of a machine language instruction that specifies the operation to be performed.