假设我们有 28 位虚拟地址空间和 32 位物理地址空间。帧大小为1kb。我想将虚拟地址0x000039A 转换为物理地址。
[我的尝试]1kb = 1024byte = 2^10 所以虚拟页码是0x0000，偏移量是0x39A。在这个页表中，0x00 有0x0100 物理页号。所以我认为物理地址号是 0x010039A 但答案是 0x0004039A 。有人能解释一下吗？
【问题讨论】：
标签： memory memory-management operating-system
让我们一步一步来，尽管here 的答案可以帮助你自己做。
__________
| STEP 1 |
¯¯¯¯¯¯¯¯¯¯
0x000039A is in virtual, lets convert it to binary representation
0b 0000 0000 0000 0000 0011 1001 1010
0x 0 0 0 0 3 9 A
__________
| STEP 2 |
¯¯¯¯¯¯¯¯¯¯
Calculate the number of bits needed to reference the whole 1KB
1K = 2^10
==> 10 bits are needed. Just do log2(page-size).
__________
| STEP 3 |
¯¯¯¯¯¯¯¯¯¯
Take away the first 10 bits of the binary presentation
0b 0000 0000 0000 0000 0011 1001 1010
offset = 0b 11 1001 1010
= 0x 3 9 A
__________
| STEP 4 |
¯¯¯¯¯¯¯¯¯¯
Get the virtual page out of what ever bits left
0b (00)(00 00)(00 00)(00 00)(00 00)
= 0x00000
__________
| STEP 5 |
¯¯¯¯¯¯¯¯¯¯
Go to the page table at the entry 0x00000, there you will find the corresponding frame number.
Suppose the page table is given:
________________
| 0x0 | 0x0100 |
| 0x1 | 0xA |
| . | |
| . | |
| | |
----------------
__________
| STEP 6 |
¯¯¯¯¯¯¯¯¯¯
Turn the frame number to binary representation and concatenate it to the offset
Frame | offset
0x0100 |
0b (00)(00 00)(01 00)(00 00)(00 | 11) (1001) (1010)
0x 004039A
【讨论】：