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+2 votes

The address space of 8086 CPU is

  1. one Megabyte

  2. 256 Kilobytes

  3. 1 K Megabytes

  4. 64 Kilobytes

asked in CO & Architecture by Veteran (59.5k points)
retagged by | 1.7k views

2 Answers

+7 votes
Best answer
in 8086 architecture there are 16 bit data lines and 20 address lines.
20 lines means 2^20 byte = 1 mega byte
answered by Veteran (55.1k points)
selected by
Here the data line is 16 bits. So does it mean that word size is of 16 bits??
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

@Arjun Sir

Here is the link to similar question (, 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??
+1 vote

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).

answered by Active (5k points)

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