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How a 64-bit process virtual address space is divided in Linux?
The Next CEO of Stack OverflowLinux Kernel logical address space organisationhow is page size determined in virtual address space?Physical Address Extension - how do virtual addresses work?How can two identical virtual addresses point to different physical addresses?How does the CPU knows which physical address is mapped to which virtual address?how to get virtual address generated by a processHow can the Linux kernel address from 8 MB to 1 GB of virtual memory in x86 systemsDoes the isolation between virtual memory address spaces of different processes not apply to privileged process and to swap?Do the virtual address spaces of all the processes have the same content in their “Kernel” parts?Disabling virtual address space randomization for a linux kernel module
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 11 hours ago
ChristopherChristopher
1532
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 11 hours ago
ChristopherChristopher
1532
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 11 hours ago
ChristopherChristopher
1532
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 11 hours ago
ChristopherChristopher
1532
asked 11 hours ago
ChristopherChristopher
1532
asked 11 hours ago
ChristopherChristopher
1532
asked 11 hours ago
ChristopherChristopher
1532
asked 11 hours ago
ChristopherChristopher
1532
1532
add a comment |
add a comment |
1 Answer
1
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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 11 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
9 hours ago
Thanks @Johan, I’ve tried to highlight this.
– Stephen Kitt
9 hours ago
add a comment |
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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 11 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
9 hours ago
Thanks @Johan, I’ve tried to highlight this.
– Stephen Kitt
9 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 11 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
9 hours ago
Thanks @Johan, I’ve tried to highlight this.
– Stephen Kitt
9 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 11 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 11 hours ago
Stephen KittStephen Kitt
179k24406484
edited 9 hours ago
answered 11 hours ago
Stephen KittStephen Kitt
179k24406484
answered 11 hours ago
Stephen KittStephen Kitt
179k24406484
answered 11 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
9 hours ago
Thanks @Johan, I’ve tried to highlight this.
– Stephen Kitt
9 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
9 hours ago
Thanks @Johan, I’ve tried to highlight this.
– Stephen Kitt
9 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
9 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
9 hours ago
Thanks @Johan, I’ve tried to highlight this.
– Stephen Kitt
9 hours ago
Thanks @Johan, I’ve tried to highlight this.
– Stephen Kitt
9 hours ago
add a comment |
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