Web» The offset address is contained in IP; let it be 95F3H. » The logical address is CS: IP, or 2500: 95F3H. » Then the physical address will be 25000 + 95F3 = 2E5F3H. » The lowest memory location of the code segment will be 25000H (25000+0000) » The highest memory location will be 34FFFH (25000+FFFF) MP, CSE, VCET WebJun 2, 2011 · 12. To really explain the concept, we have to back up to the basic idea of segments, and how the x86 uses them (in real mode). The 8086 has 20-bit addressing, but only 16-bit registers. To generate 20-bit addresses, it combines a segment with an offset. The segment has to be in a segment register (CS, DS, ES, or SS).

x86 memory segmentation - Wikipedia

WebApr 13, 2024 · ending address (a) 1000H → 10000H 1FFFFH (b) 1234H → 12340H 2233FH (c) 2300H → 23000H 32FFFH (d) E000H → E0000H EFFFFH (e) AB00H → AB000H … WebMay 22, 2010 · This results in a 20 bit address bus. As an example, if the CS register contains 1234H, and the IP register contains 5678H, then the next instruction is fetched from physical address 179B8H, which ... flair hair fashions

Solved Q1 / The Contents of the following registers are : CS - Chegg

WebDownload Specifications. Product Collection. 7th Generation Intel® Core™ i3 Processors. Code Name. Products formerly Kaby Lake. Vertical Segment. Mobile. Processor … Web1 Answer. The value of the code segment register (CS) = 3560h The value of the Instruction Pointer, holding address of the instruction = 101Ah Physical address of the instruction Physical Address = CS (Hex) x 16 + IP. The value of the stack segment register (SS) = 2067h The value of the stack pointer (SP) which is Offset = 1235h Thus, Physical ... WebMOV AL,[SI]+1234H Example: let you have DS = 0200H, CS = 0100H, IP = 0000H, AL=BEH, SI = 2000H, and you execute the instruction MOV AL,[SI]+1234H. How these instruction is executed and what are the memory mapping of it? • The starting address of your instruction is: PA = CS*10H+IP = 0100 * 10 + 0000 = 1000H. flair hair mens hats