Wang Shuang assembly is in 16, if segment address is FFFFH, offset for FFFFH, then calculate the physical address of the 10 ffefh beyond the FFFFFH (20), and this is how did that happen?

CodePudding user response:

The unit of segment address is different, not 1 but 16 (bytes), so the computation to shift to the left four (16) is multiplied by the

CodePudding user response:

Addressing at 8086/8088 more than 1 MB (20 address line) will be address wrap, so FFFF: FFFF physical addresses instead of 10 FFEF FFEF (0000: FFEF), but in the 286 +, if you enable A20, wrap does not address, over 1 MB there will be a piece of 64 k - 16 bytes accessible areas of the real mode, called the HMA

CodePudding user response:

Behind the Wang Shuang assembly can learn what you said?

CodePudding user response:

refer to the second floor early play big play nuclear response:

addressing at 8086/8088 more than 1 MB (20 address line) will be address wrap, so FFFF: FFFF physical addresses instead of 10 FFEF FFEF (0000: FFEF), but in the 286 +, if you enable A20, wrap does not address, over 1 MB there will be a piece of 64 k - 16 bytes accessible areas of the real mode, called the HMA

Behind the Wang Shuang assembly can learn what you said?

CodePudding user response:

Wang Shuang book seems to be no (not sure), but speak 32-bit assembly will be mentioned in the book of A20 can make it or not the influence of