4 votes 4 votes The address space of $8086$ CPU is one Megabyte $256$ Kilobytes $1 \;\text{K}$ Megabytes $64$ Kilobytes CO and Architecture gate1998 co-and-architecture microprocessors out-of-syllabus-now isro2008 + – Kathleen asked Sep 25, 2014 • edited Dec 4, 2022 by Lakshman Bhaiya Kathleen 4.6k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 10 votes 10 votes in 8086 architecture there are 16 bit data lines and 20 address lines. 20 lines means 2^20 byte = 1 mega byte Digvijay Pandey answered May 7, 2015 • selected Jun 9, 2015 by Arjun Digvijay Pandey comment Share Follow See all 6 Comments See all 6 6 Comments reply anchitjindal07 commented Apr 29, 2017 reply Follow Share Here the data line is 16 bits. So does it mean that word size is of 16 bits?? 2 votes 2 votes Arjun commented Apr 29, 2017 reply Follow Share yes. 2 votes 2 votes anchitjindal07 commented Apr 29, 2017 reply Follow Share 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 0 votes 0 votes anchitjindal07 commented Apr 29, 2017 reply Follow Share @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 0 votes 0 votes Arjun commented Apr 29, 2017 reply Follow Share 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. 1 votes 1 votes anchitjindal07 commented Apr 29, 2017 reply Follow Share "most current architectures are like that" means although they have word size of any length, they are byte addressable?? 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes 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). kvkumar answered Jul 1, 2016 kvkumar comment Share Follow See 1 comment See all 1 1 comment reply anchitjindal07 commented Dec 26, 2018 reply Follow Share 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 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes 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 topper98 answered Mar 23, 2020 topper98 comment Share Follow See all 0 reply Please log in or register to add a comment.