3 ms·

This is highly dependent on the video card and other PCI devices you are using, as they will often reserve huge chunks of VA space to map to device memory (and

This is highly dependent on the video card and other PCI devices you are using, as they will often reserve huge chunks of VA space to map to device memory (and furthermore, require it to be in low addresses i.e. user space), making a contiguous reservation difficult. If you want to see exactly what's happening, SysInternals VMMap (http://technet.microsoft.com/en-us/sysinternals/dd535533?ppud=4 http://technet.microsoft.com/en-us/sysinternals/dd535533?ppu...) will tell you exactly the VA space layout. Note that Linux has exactly the same problems with VA space, except they default to 3/1, instead of NT's default (but configurable) of 2/2. It seems that a quick fix to this issue would be to lower that number from 512MB to something like 128MB, though I know nothing about Go's internals.

JoshTriplett 15y ago
Getting 512MB of contiguous virtual address space does not require 512MB of contiguous free physical memory; contiguous virtual addresses can map to discontiguous physical memory.
- astrodust 15y ago
  That's what I thought. The address your user-space program uses and the actual address in memory are two different things, aren't they? Isn't this what enables the operating system to scramble the allocations it gives you to make it harder to implement a buffer-overflow attack?
  - JoshTriplett 15y ago
    > That's what I thought. The address your user-space program uses and the actual address in memory are two different things, aren't they? Right. Userspace uses virtual addresses, which get mapped through page tables to become physical addresses. Some kernel addresses get identity-mapped (meaning that the virtual address matches the physical address), while some kernel addresses go through translation as well. (Specifically, on Linux, kmalloc allocates identity-mapped addresses, while vmalloc allocates virtual addresses; vmalloc allows you to have a large virtually contiguous buffer without requiring a large physically contiguous region of free memory.) > Isn't this what enables the operating system to scramble the allocations it gives you to make it harder to implement a buffer-overflow attack? You can do Address Space Layout Randomization (ASLR) for either virtual or physical address spaces; you want to do it for whatever type of address an attacker could otherwise make use of. Userspace processes can use ASLR for their virtual address space, so that attackers can't make use of fixed addresses. The Linux kernel doesn't normally map virtual pages to any particular well-known location in physical address space, either. The kernel can also use ASLR for some (though not all) of its own kernel-space addresses. Beyond that, recent versions of Linux also try to avoid exposing kernel-space addresses to non-root users. Also note that ASLR doesn't generally introduce enough randomness in a 32-bit address space.
4ad 15y ago
I/O devices are never mapped in user space in Windows, and I think that's true for other operating systems as well.
- xpaulbettsx 15y ago
  Ah, you're right, they have to be in low physical frames, I'm getting confused.

JoshTriplett 15y ago

Getting 512MB of contiguous virtual address space does not require 512MB of contiguous free physical memory; contiguous virtual addresses can map to discontiguous physical memory.

astrodust 15y ago
That's what I thought. The address your user-space program uses and the actual address in memory are two different things, aren't they? Isn't this what enables the operating system to scramble the allocations it gives you to make it harder to implement a buffer-overflow attack?
- JoshTriplett 15y ago
  > That's what I thought. The address your user-space program uses and the actual address in memory are two different things, aren't they? Right. Userspace uses virtual addresses, which get mapped through page tables to become physical addresses. Some kernel addresses get identity-mapped (meaning that the virtual address matches the physical address), while some kernel addresses go through translation as well. (Specifically, on Linux, kmalloc allocates identity-mapped addresses, while vmalloc allocates virtual addresses; vmalloc allows you to have a large virtually contiguous buffer without requiring a large physically contiguous region of free memory.) > Isn't this what enables the operating system to scramble the allocations it gives you to make it harder to implement a buffer-overflow attack? You can do Address Space Layout Randomization (ASLR) for either virtual or physical address spaces; you want to do it for whatever type of address an attacker could otherwise make use of. Userspace processes can use ASLR for their virtual address space, so that attackers can't make use of fixed addresses. The Linux kernel doesn't normally map virtual pages to any particular well-known location in physical address space, either. The kernel can also use ASLR for some (though not all) of its own kernel-space addresses. Beyond that, recent versions of Linux also try to avoid exposing kernel-space addresses to non-root users. Also note that ASLR doesn't generally introduce enough randomness in a 32-bit address space.

4ad 15y ago

I/O devices are never mapped in user space in Windows, and I think that's true for other operating systems as well.

xpaulbettsx 15y ago
Ah, you're right, they have to be in low physical frames, I'm getting confused.