Effective Address Calculation for 20(R1) and -(R2) Addressing Modes

To find an effective address (EA), you apply the rule for each addressing mode to the contents of the specified register.

Key idea: the displacement (e.g., 20) is added to (or used with) a base register, while pre-decrement (e.g., -(R2)) first modifies the register and then uses the updated value as the EA.

We’ll compute:

• (i) Load 20(R1), R5
• (ii) Add –(R2), R5

Let:

• $R1 = 1200$
• $R2 = 4600$

Addressing modes involved:

• Register-Relative Addressing d(Rbase)
• Displacement d
• Predecrement Addressing -(R)
• Effective Address EA

(i) Load 20(R1), R5 → EA calculation

For register-relative (base + displacement) addressing of the form d(R1), the effective address is:

Here, d = 20 and R1 = 1200, so:

Therefore, the memory operand in (i) is located at effective address 1220.

Register-Relative Addressing is exactly what this form represents: base register value plus displacement.

(ii) Add –(R2), R5 → EA calculation

The notation -(R2) indicates pre-decrement addressing:

So the effective address depends on the decrement amount, which is typically the size of the operand (e.g., 1, 2, 4 bytes) in the target architecture / instruction set.

Most textbook problems implicitly assume a decrement of 1 memory unit when the operand-size detail is not provided. Under that common assumption:

1. Decrement $R2$ first: $$R2 \leftarrow 4600 - 1 = 4599$$
2. Use the updated $R2$ as the effective address: EA = 4599 $$

Therefore, under the standard assumption of decrement-by-1: the memory operand in (ii) is located at effective address 4599.

If your course specifies operand size (e.g., byte/word/long), replace the “1” with the architecture’s unit decrement (e.g., 2 or 4). The process stays the same; only the decrement amount changes.