GATE CSE 1998 | Question: 2.10, ISRO2008-17

Question

The address space of $8086$ CPU is

• A. one Megabyte

• B. $256$ Kilobytes

• C. $1 \;\text{K}$ Megabytes

• D. $64$ Kilobytes

Answer 1

in 8086 architecture there are 16 bit data lines and 20 address lines.
20 lines means 2^20 byte = 1 mega byte

Comments

@Arjun Sir

Here is the link to similar question (https://gateoverflow.in/42603/what-will-size-memory-address-space-bit-data-and-bit-address), where you have selected 2MB as the correct answer. But in this question answer is 1 MB. Please tell which one is correct. If both are correct then what is the difference between these two

---

you can read the comments there. In this question it says "8086" and it is byte addressable even though word size is 16 bits. Actually most current architectures are like that.

---

"most current architectures are like that" means although they have word size of any length, they are byte addressable??

Answer 2

All internal registers, as well as internal and external data buses, are 16 bits wide, which firmly established the "16-bit microprocessor" identity of the 8086. A 20-bit external address bus provides a 1 MB physicaladdress space (2²⁰ = 1,048,576).

Comments

Sir if word size is 16 bits (=2 bytes), then should not the answer to the question be 2 MB? Here it has 20 bit address line, so it can address 2^20 words each of whose size is 2 bytes. So Address space should be 2*2^20 Bytes which is equal to 2 MB. Please tell where I am wrong

Answer 3

In 8086 architecture there are 16 bit data lines and 20 bit address lines.
16 bit data lines means that the word size must be 2 bytes and these 2 bytes can be read in a single memory cycle.
20 bit address lines corresponds that the memory size is 220 Bytes = 1 Megabyte