How a 64-bit process virtual address space is divided in Linux?
The following image shows how a 32-bit process virtual address space is divided:
But how a 64-bit process virtual address space is divided?
linux
asked 4 hours ago
ChristopherChristopher
1382
add a comment |
The following image shows how a 32-bit process virtual address space is divided:
But how a 64-bit process virtual address space is divided?
linux
asked 4 hours ago
ChristopherChristopher
1382
add a comment |
The following image shows how a 32-bit process virtual address space is divided:
But how a 64-bit process virtual address space is divided?
linux
asked 4 hours ago
ChristopherChristopher
1382
The following image shows how a 32-bit process virtual address space is divided:
But how a 64-bit process virtual address space is divided?
linux
linux
asked 4 hours ago
ChristopherChristopher
1382
asked 4 hours ago
ChristopherChristopher
1382
asked 4 hours ago
ChristopherChristopher
1382
asked 4 hours ago
ChristopherChristopher
1382
asked 4 hours ago
ChristopherChristopher
1382
1382
add a comment |
add a comment |
1 Answer
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The 64-bit x86 virtual memory map splits the address space into two: the lower section (with the top bit set to 0) is user-space, the upper section (with the top bit set to 1) is kernel-space. (Note that x86-64 defines “canonical” “lower half” and “higher half” addresses, with a number of bits effectively limited to 48 or 56; see Wikipedia for details.)
The complete map is documented in detail in the kernel; currently it looks like
===========================================================================================
Start addr | Offset | End addr | Size | VM area description
===========================================================================================
| | | |
0000000000000000 | 0 | 00007fffffffffff | 128 TB | user-space virtual memory
__________________|____________|__________________|_________|______________________________
| | | |
0000800000000000 | +128 TB | ffff7fffffffffff | ~16M TB | non-canonical
__________________|____________|__________________|_________|______________________________
| | | |
ffff800000000000 | -128 TB | ffffffffffffffff | 128 TB | kernel-space virtual memory
__________________|____________|__________________|_________|______________________________
with 48-bit virtual addresses.
Unlike the 32-bit case, the “64-bit” memory map is a direct reflection of hardware constraints.
answered 3 hours ago
Stephen KittStephen Kitt
179k24406484
To clarify: this limitation is imposed by the hardware. There is currently no 64-bit processor implementation that doesn't leave a huge hole of unusable addresses in the middle of the virtual address space. The amount of physical memory the CPUs are able to address is also way below 2 to the power of 64.
– Johan Myréen
2 hours ago
Thanks @Johan, I’ve tried to highlight this.
– Stephen Kitt
2 hours ago
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
The 64-bit x86 virtual memory map splits the address space into two: the lower section (with the top bit set to 0) is user-space, the upper section (with the top bit set to 1) is kernel-space. (Note that x86-64 defines “canonical” “lower half” and “higher half” addresses, with a number of bits effectively limited to 48 or 56; see Wikipedia for details.)
The complete map is documented in detail in the kernel; currently it looks like
===========================================================================================
Start addr | Offset | End addr | Size | VM area description
===========================================================================================
| | | |
0000000000000000 | 0 | 00007fffffffffff | 128 TB | user-space virtual memory
__________________|____________|__________________|_________|______________________________
| | | |
0000800000000000 | +128 TB | ffff7fffffffffff | ~16M TB | non-canonical
__________________|____________|__________________|_________|______________________________
| | | |
ffff800000000000 | -128 TB | ffffffffffffffff | 128 TB | kernel-space virtual memory
__________________|____________|__________________|_________|______________________________
with 48-bit virtual addresses.
Unlike the 32-bit case, the “64-bit” memory map is a direct reflection of hardware constraints.
answered 3 hours ago
Stephen KittStephen Kitt
179k24406484
To clarify: this limitation is imposed by the hardware. There is currently no 64-bit processor implementation that doesn't leave a huge hole of unusable addresses in the middle of the virtual address space. The amount of physical memory the CPUs are able to address is also way below 2 to the power of 64.
– Johan Myréen
2 hours ago
Thanks @Johan, I’ve tried to highlight this.
– Stephen Kitt
2 hours ago
add a comment |
The 64-bit x86 virtual memory map splits the address space into two: the lower section (with the top bit set to 0) is user-space, the upper section (with the top bit set to 1) is kernel-space. (Note that x86-64 defines “canonical” “lower half” and “higher half” addresses, with a number of bits effectively limited to 48 or 56; see Wikipedia for details.)
The complete map is documented in detail in the kernel; currently it looks like
===========================================================================================
Start addr | Offset | End addr | Size | VM area description
===========================================================================================
| | | |
0000000000000000 | 0 | 00007fffffffffff | 128 TB | user-space virtual memory
__________________|____________|__________________|_________|______________________________
| | | |
0000800000000000 | +128 TB | ffff7fffffffffff | ~16M TB | non-canonical
__________________|____________|__________________|_________|______________________________
| | | |
ffff800000000000 | -128 TB | ffffffffffffffff | 128 TB | kernel-space virtual memory
__________________|____________|__________________|_________|______________________________
with 48-bit virtual addresses.
Unlike the 32-bit case, the “64-bit” memory map is a direct reflection of hardware constraints.
answered 3 hours ago
Stephen KittStephen Kitt
179k24406484
To clarify: this limitation is imposed by the hardware. There is currently no 64-bit processor implementation that doesn't leave a huge hole of unusable addresses in the middle of the virtual address space. The amount of physical memory the CPUs are able to address is also way below 2 to the power of 64.
– Johan Myréen
2 hours ago
Thanks @Johan, I’ve tried to highlight this.
– Stephen Kitt
2 hours ago
add a comment |
The 64-bit x86 virtual memory map splits the address space into two: the lower section (with the top bit set to 0) is user-space, the upper section (with the top bit set to 1) is kernel-space. (Note that x86-64 defines “canonical” “lower half” and “higher half” addresses, with a number of bits effectively limited to 48 or 56; see Wikipedia for details.)
The complete map is documented in detail in the kernel; currently it looks like
===========================================================================================
Start addr | Offset | End addr | Size | VM area description
===========================================================================================
| | | |
0000000000000000 | 0 | 00007fffffffffff | 128 TB | user-space virtual memory
__________________|____________|__________________|_________|______________________________
| | | |
0000800000000000 | +128 TB | ffff7fffffffffff | ~16M TB | non-canonical
__________________|____________|__________________|_________|______________________________
| | | |
ffff800000000000 | -128 TB | ffffffffffffffff | 128 TB | kernel-space virtual memory
__________________|____________|__________________|_________|______________________________
with 48-bit virtual addresses.
Unlike the 32-bit case, the “64-bit” memory map is a direct reflection of hardware constraints.
answered 3 hours ago
Stephen KittStephen Kitt
179k24406484
The 64-bit x86 virtual memory map splits the address space into two: the lower section (with the top bit set to 0) is user-space, the upper section (with the top bit set to 1) is kernel-space. (Note that x86-64 defines “canonical” “lower half” and “higher half” addresses, with a number of bits effectively limited to 48 or 56; see Wikipedia for details.)
The complete map is documented in detail in the kernel; currently it looks like
===========================================================================================
Start addr | Offset | End addr | Size | VM area description
===========================================================================================
| | | |
0000000000000000 | 0 | 00007fffffffffff | 128 TB | user-space virtual memory
__________________|____________|__________________|_________|______________________________
| | | |
0000800000000000 | +128 TB | ffff7fffffffffff | ~16M TB | non-canonical
__________________|____________|__________________|_________|______________________________
| | | |
ffff800000000000 | -128 TB | ffffffffffffffff | 128 TB | kernel-space virtual memory
__________________|____________|__________________|_________|______________________________
with 48-bit virtual addresses.
Unlike the 32-bit case, the “64-bit” memory map is a direct reflection of hardware constraints.
answered 3 hours ago
Stephen KittStephen Kitt
179k24406484
edited 2 hours ago
answered 3 hours ago
Stephen KittStephen Kitt
179k24406484
answered 3 hours ago
Stephen KittStephen Kitt
179k24406484
answered 3 hours ago
Stephen KittStephen Kitt
179k24406484
179k24406484
To clarify: this limitation is imposed by the hardware. There is currently no 64-bit processor implementation that doesn't leave a huge hole of unusable addresses in the middle of the virtual address space. The amount of physical memory the CPUs are able to address is also way below 2 to the power of 64.
– Johan Myréen
2 hours ago
Thanks @Johan, I’ve tried to highlight this.
– Stephen Kitt
2 hours ago
add a comment |
To clarify: this limitation is imposed by the hardware. There is currently no 64-bit processor implementation that doesn't leave a huge hole of unusable addresses in the middle of the virtual address space. The amount of physical memory the CPUs are able to address is also way below 2 to the power of 64.
– Johan Myréen
2 hours ago
Thanks @Johan, I’ve tried to highlight this.
– Stephen Kitt
2 hours ago
To clarify: this limitation is imposed by the hardware. There is currently no 64-bit processor implementation that doesn't leave a huge hole of unusable addresses in the middle of the virtual address space. The amount of physical memory the CPUs are able to address is also way below 2 to the power of 64.
– Johan Myréen
2 hours ago
To clarify: this limitation is imposed by the hardware. There is currently no 64-bit processor implementation that doesn't leave a huge hole of unusable addresses in the middle of the virtual address space. The amount of physical memory the CPUs are able to address is also way below 2 to the power of 64.
– Johan Myréen
2 hours ago
Thanks @Johan, I’ve tried to highlight this.
– Stephen Kitt
2 hours ago
Thanks @Johan, I’ve tried to highlight this.
– Stephen Kitt
2 hours ago
add a comment |
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