Calculation of 8086-segment offset

Preface

80,861 Total 20 address lines, addressing the physical address with a 16bit segment address + 4bit offset.
The same physical address can represent a test program with multiple logical addresses (different segment addresses + offsets)

Testcase1.cpp:Defines the entry point for the console application. #include "stdafx.h" #include <stdlib.h> #include <stdio.h> #include <assert.h>//Physical address based on segment and offset long

GET8086PHADDR (/*in*/int isegment,/*in*/int ioffset);

Get the Segment address int get8086segment (/*in*/long lphaddr,/*in*/int ioffset) According to the physical address and offset;

Get offset int Get8086offset (/*in*/long lphaddr,/*in*/int isegment) According to physical address and segment address;
    int main (int argc, char* argv[]) {long ltmp = 0;
    Long lphaddr = 0;
    int isegment = 0;

    int ioffset = 0;
    -D 123:456 lphaddr = Get8086phaddr (0x123, 0x456);

    printf ("0x%X = 123:456\n", lphaddr); -d?
    : 186 isegment = Get8086segment (lphaddr, 0x186);
    Ltmp = Get8086phaddr (isegment, 0x186);
    ASSERT (lphaddr = = ltmp);

    printf ("0x%X =%x:186\n", ltmp, isegment);
    -D 105:?
    Ioffset = Get8086offset (lphaddr, 0x105);
    Ltmp = Get8086phaddr (0x105, Ioffset);
    ASSERT (lphaddr = = ltmp);

  printf ("0x%X = 105:%x\n", ltmp, Ioffset);  /* Run result 0x1686 = 123:456 0x1686 = 150:186 0x1686 = 105:636 Press any key to continue ... With debug authentication, the memory results are the same in the following way, indicating that a physical address can be represented by multiple logical addresses------123:456-d 150:186-d 105:636-d 0:1686 */System ("p
    Ause ");
return 0;

} long get8086phaddr (/*in*/int isegment,/*in*/int ioffset) {return ((long) isegment << 4) + Ioffset;}

int get8086segment (/*in*/long lphaddr,/*in*/int ioffset) {return ((Lphaddr-ioffset) >> 4);}

 int Get8086offset (/*in*/long lphaddr,/*in*/int isegment) {return lphaddr-((Long) isegment << 4);}

Validation