Question

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，因此虚拟页码为0x0000，偏移量为0x39A。在此页表中，0x00具有0x0100物理页码。因此，我认为物理地址为0x010039A，但答案为0x0004039A。有人可以解释吗？

Answer 1

尽管答案here可以帮助您完成自己的工作，但还是要逐步进行。

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| STEP 1 |
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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

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| STEP 2 |
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Calculate the number of bits needed to reference the whole 1KB

1K = 2^10

==> 10 bits are needed. Just do log2(page-size).

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| STEP 3 |
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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 

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| STEP 4 |
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Get the virtual page out of what ever bits left

0b (00)(00 00)(00 00)(00 00)(00 00)
=  0x00000

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| STEP 5 |
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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    |
|  .  |        |
|  .  |        |
|     |        |
----------------

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| STEP 6 |
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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

将虚拟地址转换为物理地址

1 个答案: