Virtual address to physical address translation (in decimal)What is private bytes, virtual bytes, working set?Mapping of Virtual Address to Physical AddressWhat are the differences between virtual memory and physical memory?converting a logical address to a physical addressVirtual Address to Physical addressProcessors and virtual/physical addressesTranslating from logical to physical addressestranslate virtual address to physical addressConvert virtual address to physical addressConverting virtual to physical address
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Virtual address to physical address translation (in decimal)
What is private bytes, virtual bytes, working set?Mapping of Virtual Address to Physical AddressWhat are the differences between virtual memory and physical memory?converting a logical address to a physical addressVirtual Address to Physical addressProcessors and virtual/physical addressesTranslating from logical to physical addressestranslate virtual address to physical addressConvert virtual address to physical addressConverting virtual to physical address
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Assume a system with 6 bit virtual address and 16 byte pages per frame. The mapping of virtual page numbers to physical page of a process is:
Virtual Page Physical Page
0 8
1 3
2 11
3 1
Translate the virtual address 40 (which is in decimal) to physical address in decimal.
From the 6 bit virtual address, I know there are 2^6 addresses in the virtual address space.
And there are 2^4 bytes pages per frame.
So there are (2^6)/(2^4) = 2^2 bytes to each address page and that means there is a 4 bit offset.
However, I do not know where to go on from there and use the table to translate the virtual address to physical address in decimal.
operating-system virtual
edited Mar 8 at 0:20
Remy Lebeau
342k19267461
asked Mar 8 at 0:18
encryptencrypt
61
add a comment |
Assume a system with 6 bit virtual address and 16 byte pages per frame. The mapping of virtual page numbers to physical page of a process is:
Virtual Page Physical Page
0 8
1 3
2 11
3 1
Translate the virtual address 40 (which is in decimal) to physical address in decimal.
From the 6 bit virtual address, I know there are 2^6 addresses in the virtual address space.
And there are 2^4 bytes pages per frame.
So there are (2^6)/(2^4) = 2^2 bytes to each address page and that means there is a 4 bit offset.
However, I do not know where to go on from there and use the table to translate the virtual address to physical address in decimal.
operating-system virtual
edited Mar 8 at 0:20
Remy Lebeau
342k19267461
asked Mar 8 at 0:18
encryptencrypt
61
add a comment |
Assume a system with 6 bit virtual address and 16 byte pages per frame. The mapping of virtual page numbers to physical page of a process is:
Virtual Page Physical Page
0 8
1 3
2 11
3 1
Translate the virtual address 40 (which is in decimal) to physical address in decimal.
From the 6 bit virtual address, I know there are 2^6 addresses in the virtual address space.
And there are 2^4 bytes pages per frame.
So there are (2^6)/(2^4) = 2^2 bytes to each address page and that means there is a 4 bit offset.
However, I do not know where to go on from there and use the table to translate the virtual address to physical address in decimal.
operating-system virtual
edited Mar 8 at 0:20
Remy Lebeau
342k19267461
asked Mar 8 at 0:18
encryptencrypt
61
Assume a system with 6 bit virtual address and 16 byte pages per frame. The mapping of virtual page numbers to physical page of a process is:
Virtual Page Physical Page
0 8
1 3
2 11
3 1
Translate the virtual address 40 (which is in decimal) to physical address in decimal.
From the 6 bit virtual address, I know there are 2^6 addresses in the virtual address space.
And there are 2^4 bytes pages per frame.
So there are (2^6)/(2^4) = 2^2 bytes to each address page and that means there is a 4 bit offset.
However, I do not know where to go on from there and use the table to translate the virtual address to physical address in decimal.
operating-system virtual
operating-system virtual
edited Mar 8 at 0:20
Remy Lebeau
342k19267461
asked Mar 8 at 0:18
encryptencrypt
61
edited Mar 8 at 0:20
Remy Lebeau
342k19267461
asked Mar 8 at 0:18
encryptencrypt
61
edited Mar 8 at 0:20
Remy Lebeau
342k19267461
edited Mar 8 at 0:20
Remy Lebeau
342k19267461
edited Mar 8 at 0:20
Remy Lebeau
342k19267461
342k19267461
asked Mar 8 at 0:18
encryptencrypt
61
asked Mar 8 at 0:18
encryptencrypt
61
asked Mar 8 at 0:18
encryptencrypt
61
61
add a comment |
add a comment |
1 Answer
1
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40=0000000101000b
If we interpret this address for the VM 000.000010.1000b.
offset=1000b (4LSB)
virtual page number=000010 (6 next bits)
So virtual page number=2. If we look at the page table, we see that virtual page 2 maps to physical page 11=001011b
Hence physical address will be 00.001011.1000=8+16+32+128=184d
answered Mar 8 at 9:00
Alain MerigotAlain Merigot
4,4931822
add a comment |
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40=0000000101000b
If we interpret this address for the VM 000.000010.1000b.
offset=1000b (4LSB)
virtual page number=000010 (6 next bits)
So virtual page number=2. If we look at the page table, we see that virtual page 2 maps to physical page 11=001011b
Hence physical address will be 00.001011.1000=8+16+32+128=184d
answered Mar 8 at 9:00
Alain MerigotAlain Merigot
4,4931822
add a comment |
40=0000000101000b
If we interpret this address for the VM 000.000010.1000b.
offset=1000b (4LSB)
virtual page number=000010 (6 next bits)
So virtual page number=2. If we look at the page table, we see that virtual page 2 maps to physical page 11=001011b
Hence physical address will be 00.001011.1000=8+16+32+128=184d
answered Mar 8 at 9:00
Alain MerigotAlain Merigot
4,4931822
add a comment |
40=0000000101000b
If we interpret this address for the VM 000.000010.1000b.
offset=1000b (4LSB)
virtual page number=000010 (6 next bits)
So virtual page number=2. If we look at the page table, we see that virtual page 2 maps to physical page 11=001011b
Hence physical address will be 00.001011.1000=8+16+32+128=184d
answered Mar 8 at 9:00
Alain MerigotAlain Merigot
4,4931822
40=0000000101000b
If we interpret this address for the VM 000.000010.1000b.
offset=1000b (4LSB)
virtual page number=000010 (6 next bits)
So virtual page number=2. If we look at the page table, we see that virtual page 2 maps to physical page 11=001011b
Hence physical address will be 00.001011.1000=8+16+32+128=184d
answered Mar 8 at 9:00
Alain MerigotAlain Merigot
4,4931822
answered Mar 8 at 9:00
Alain MerigotAlain Merigot
4,4931822
answered Mar 8 at 9:00
Alain MerigotAlain Merigot
4,4931822
answered Mar 8 at 9:00
Alain MerigotAlain Merigot
4,4931822
4,4931822
add a comment |
add a comment |
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