最快的Linux系统调用 [英] Fastest Linux system call

问题描述

在支持syscall和sysret的x86-64 Intel系统上，香草内核上来自64位用户代码的最快"系统调用是什么?

On an x86-64 Intel system that supports syscall and sysret what's the "fastest" system call from 64-bit user code on a vanilla kernel?

特别是，它必须是一个行使syscall/sysret用户<->内核转换 ¹ 的系统调用，但执行的工作量最少.它甚至不需要执行syscall本身:只要不因某些原因而走慢路，就可以从不分派给内核方面的特定调用的某种类型的早期错误是可以的.

In particular, it must be a system call that exercises the syscall/sysret user <-> kernel transition¹, but does the least amount of work beyond that. It doesn't even need to do the syscall itself: some type of early error which never dispatches to the specific call on the kernel side is fine, as long as it doesn't go down some slow path because of that.

这样的调用可以用来估计原始的syscall和sysret开销，而与该调用所做的任何工作无关.

Such a call could be used to estimate the raw syscall and sysret overhead independent of any work done by the call.

¹ 特别是，这排除了看似是系统调用但在VDSO中实现的内容(例如，clock_gettime)或在运行时被缓存的内容(例如，getpid).

¹ In particular, this excludes things that appear to be system calls but are implemented in the VDSO (e.g., clock_gettime) or are cached by the runtime (e.g., getpid).

推荐答案

一个不存在的，因此会迅速返回-ENOSYS.

One that doesn't exist, and therefore returns -ENOSYS quickly.

从arch/x86/entry/entry_64.S:

From arch/x86/entry/entry_64.S:

#if __SYSCALL_MASK == ~0
    cmpq    $__NR_syscall_max, %rax
#else
    andl    $__SYSCALL_MASK, %eax
    cmpl    $__NR_syscall_max, %eax
#endif
    ja  1f              /* return -ENOSYS (already in pt_regs->ax) */
    movq    %r10, %rcx

    /*
     * This call instruction is handled specially in stub_ptregs_64.
     * It might end up jumping to the slow path.  If it jumps, RAX
     * and all argument registers are clobbered.
     */
#ifdef CONFIG_RETPOLINE
    movq    sys_call_table(, %rax, 8), %rax
    call    __x86_indirect_thunk_rax
#else
    call    *sys_call_table(, %rax, 8)
#endif
.Lentry_SYSCALL_64_after_fastpath_call:

    movq    %rax, RAX(%rsp)
1:

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