SchOS/kernel/dev/dbg.c

// SPDX-License-Identifier: Apache-2.0
//
// /dev/dbg -- asking the kernel a question. See include/dbg.h.
//
// The device is a conversation: write a line, read the answer.
//
//     # echo dma > /dev/dbg && cat /dev/dbg
//
// That shape is chosen so the debugger is usable with the tools already on the
// machine, before `schdbg` is built and on a system where it will not start.
// docs/plan/debugger.md argues the case at length; the short version is that a
// protocol you cannot drive by hand is a protocol you cannot debug.
//
// One conversation at a time, deliberately. The answer buffer is global and
// the lock is held only while it is being filled, so two programs asking at
// once get one answer each, in some order, and neither gets half of the
// other's. Making it per-open would mean a struct file hook for something that
// has exactly one user.

#include "types.h"
#include "defs.h"
#include "param.h"
#include "spinlock.h"
#include "sleeplock.h"
#include "fs.h"
#include "file.h"
#include "errno.h"
#include "sbuf.h"
#include "memlayout.h"
#include "mmu.h"
#include "dbg.h"

#define NVIEW    24
#define DBGBUF   (16 * 1024)

struct view {
  const char *name;
  dbg_view_fn fn;
  const char *what;
};

static struct view views[NVIEW];
static int nview;

struct claimant {
  const char *name;
  dbg_whose_fn fn;
};
static struct claimant claimants[NVIEW];
static int nclaim;

static int streq(const char *a, const char *b);

// Static, and 16 KB of it. A view of every process on a 64-slot table is the
// largest thing here and comes to a few kilobytes; the alternative -- kalloc
// on every read -- would put an allocation on the path of the tool you reach
// for when the allocator is what you are investigating.
static char        dbgbuf[DBGBUF];
static int         dbglen;
static struct spinlock dbglock;
static int         dbginit;

void
dbg_register(const char *name, dbg_view_fn fn, const char *what)
{
  if (nview >= NVIEW || !name || !fn)
    return;
  // Registering the same name twice is a bug in the caller -- it means an init
  // that runs once per CPU, which is how `watch` first appeared in the list
  // twice. Refusing keeps the list honest whoever makes that mistake next.
  for (int i = 0; i < nview; i++)
    if (streq(views[i].name, name))
      return;
  views[nview].name = name;
  views[nview].fn   = fn;
  views[nview].what = what ? what : "";
  nview++;
}

void
dbg_register_whose(const char *name, dbg_whose_fn fn)
{
  if (nclaim >= NVIEW || !name || !fn)
    return;
  for (int i = 0; i < nclaim; i++)
    if (streq(claimants[i].name, name))
      return;
  claimants[nclaim].name = name;
  claimants[nclaim].fn   = fn;
  nclaim++;
}

static int
streq(const char *a, const char *b)
{
  while (*a && *b && *a == *b) { a++; b++; }
  return *a == 0 && *b == 0;
}

static void
do_help(struct sbuf *out)
{
  sb_puts(out, "views this kernel publishes:\n");
  for (int i = 0; i < nview; i++) {
    sb_puts(out, "  ");
    sb_pad(out, views[i].name, 10);
    sb_puts(out, views[i].what);
    sb_putc(out, '\n');
  }
  sb_puts(out, "\n  whose <addr>   every subsystem's claim on one page\n");
  sb_puts(out, "  all            every view above\n");
  sb_puts(out, "  help           this\n");
}

// "0x3eceb000", or "3eceb000", or a decimal. Returns 0 if there was no number.
static int
parse_addr(const char *s, uint64 *out)
{
  uint64 v = 0;
  int    any = 0, hex = 0;

  while (*s == ' ')
    s++;
  if (s[0] == '0' && (s[1] == 'x' || s[1] == 'X')) { hex = 1; s += 2; }

  // A bare string of hex digits is taken as hex: every address in this
  // kernel's own output is printed without the 0x, and retyping one and
  // getting a decimal reading of it would be a trap.
  if (!hex) {
    const char *p = s;
    while (*p) {
      if ((*p >= 'a' && *p <= 'f') || (*p >= 'A' && *p <= 'F')) { hex = 1; break; }
      if (!(*p >= '0' && *p <= '9')) break;
      p++;
    }
    if (!hex) {
      // All decimal digits, but an address that long is hex in practice.
      int n = 0;
      const char *q = s;
      while (*q >= '0' && *q <= '9') { n++; q++; }
      if (n > 10) hex = 1;
    }
  }

  while (*s) {
    int d;
    if (*s >= '0' && *s <= '9')      d = *s - '0';
    else if (*s >= 'a' && *s <= 'f') d = *s - 'a' + 10;
    else if (*s >= 'A' && *s <= 'F') d = *s - 'A' + 10;
    else break;
    if (!hex && d > 9)
      break;
    v = hex ? (v << 4) + d : v * 10 + d;
    any = 1;
    s++;
  }

  *out = v;
  return any;
}

static void
do_whose(struct sbuf *out, const char *arg)
{
  struct whose w;
  uint64 addr, kva;

  if (!parse_addr(arg, &addr)) {
    sb_puts(out, "whose <addr>  -- a physical address, or a kernel virtual one\n");
    return;
  }

  // Which kind of address is it? Anything at or above KERNBASE is a kernel
  // virtual address and is translated; anything below is taken as physical.
  //
  // A *user* virtual address cannot be resolved here and is not guessed at:
  // it means nothing without saying which address space, and answering for
  // whichever one happens to be current would be wrong in a way that looks
  // right. `vma <pid>` is the view for that question.
  if (addr >= (uint64)KERNBASE) {
    kva  = addr;
    addr = addr - (uint64)KERNBASE;
  } else {
    kva = addr + (uint64)KERNBASE;
  }

  w.phys        = addr & ~(uint64)(PGSIZE - 1);
  w.off         = addr & (PGSIZE - 1);
  w.owners      = 0;
  w.free_owners = 0;
  w.mappings    = 0;

  sb_puts(out, "phys 0x");
  sb_puthex(out, w.phys, 16);
  if (w.off) {
    sb_puts(out, " + 0x");
    sb_puthex(out, w.off, 3);
  }
  sb_puts(out, "   kernel va 0x");
  sb_puthex(out, (kva & ~(uint64)(PGSIZE - 1)), 16);
  sb_putc(out, '\n');

  if (w.phys >= (uint64)PHYSTOP)
    sb_puts(out, "  (above PHYSTOP -- this is not RAM this kernel manages)\n");

  for (int i = 0; i < nclaim; i++)
    claimants[i].fn(out, &w);

  // The verdict. Two shapes of wrong are worth naming, because both are
  // silent everywhere else and both corrupt memory some distance away.
  sb_putc(out, '\n');
  if (w.owners == 0 && w.mappings == 0) {
    sb_puts(out, "  nobody claims this page\n");
  } else if (w.owners > 1) {
    sb_puts(out, "  *** ");
    sb_putu(out, w.owners);
    sb_puts(out, " owners -- this page is claimed more than once, and whichever\n"
                 "      allocator hands it out second will be handing out memory\n"
                 "      the first one is still using\n");
  } else if (w.free_owners > 0 && w.mappings > 0) {
    sb_puts(out, "  *** its allocator believes this page is free, and ");
    sb_putu(out, w.mappings);
    sb_puts(out, " page table(s)\n"
                 "      still map it -- it will be handed out from under them\n");
  } else if (w.owners == 0) {
    // Not a finding. kalloc keeps a free list and no record of what it has
    // handed out, so a page in use by the kernel and a page genuinely lost
    // look the same from here -- and saying "nobody owns this" about an
    // ordinary allocated page would be the query crying wolf.
    sb_puts(out, "  no allocator has a record of this page, which is what an\n"
                 "      allocated one looks like: kalloc tracks only what is free.\n");
    if (w.mappings) {
      sb_puts(out, "      ");
      sb_putu(out, w.mappings);
      sb_puts(out, " page table(s) map it, so it is in use.\n");
    }
  } else {
    sb_puts(out, "  consistent: ");
    sb_putu(out, w.owners);
    sb_puts(out, " owner, ");
    sb_putu(out, w.mappings);
    sb_puts(out, " mapping(s)\n");
  }
}

static void
do_all(struct sbuf *out)
{
  for (int i = 0; i < nview; i++) {
    sb_puts(out, "=== ");
    sb_puts(out, views[i].name);
    sb_puts(out, " ===\n");
    views[i].fn(out, "");
    sb_putc(out, '\n');
  }
}

int
dbg_command(const char *line, struct sbuf *out)
{
  char word[32], arg[64];
  int  n = 0;

  while (*line == ' ' || *line == '\t')
    line++;
  while (*line && *line != '\n' && *line != ' ' && n < (int)sizeof(word) - 1)
    word[n++] = *line++;
  word[n] = 0;

  while (*line == ' ' || *line == '\t')
    line++;
  n = 0;
  while (*line && *line != '\n' && n < (int)sizeof(arg) - 1)
    arg[n++] = *line++;
  while (n > 0 && (arg[n-1] == ' ' || arg[n-1] == '\r'))
    n--;
  arg[n] = 0;

  if (word[0] == 0 || streq(word, "help")) {
    do_help(out);
    return out->len;
  }
  if (streq(word, "all")) {
    do_all(out);
    return out->len;
  }
  if (streq(word, "whose")) {
    do_whose(out, arg);
    return out->len;
  }

  for (int i = 0; i < nview; i++) {
    if (streq(word, views[i].name)) {
      views[i].fn(out, arg);
      return out->len;
    }
  }

  sb_puts(out, "no view called '");
  sb_puts(out, word);
  sb_puts(out, "'. try: help\n");
  return out->len;
}

// Fill dbgbuf, and say so if the answer did not fit.
//
// sb_* silently drops what will not fit, which is the right behaviour for a
// builder that must not fail -- but a table that stops in the middle of a row
// with no explanation reads as a kernel that is broken rather than as a buffer
// that is full. The last 64 bytes are held back to have somewhere to say it.
static int
fill(const char *cmd)
{
  struct sbuf sb;

  sb.p = dbgbuf; sb.len = 0; sb.cap = DBGBUF - 64;
  if (cmd)
    dbg_command(cmd, &sb);
  else
    do_all(&sb);

  if (sb.len >= sb.cap - 1) {
    sb.cap = DBGBUF;
    sb_puts(&sb, "\n-- truncated: the answer is larger than the buffer. "
                 "ask for one view at a time.\n");
  }
  dbgbuf[sb.len] = 0;
  return sb.len;
}

void
dbg_dump_all(void)
{
  if (!dbginit) { initlock(&dbglock, "dbg"); dbginit = 1; }

  acquire(&dbglock);
  dbglen = fill(0);
  // Printed inside the lock so a second sysrq cannot interleave with this one.
  // cprintf takes cons.lock, which is never held by anything that wants
  // dbglock, so the order is fixed and there is no cycle.
  cprintf("%s", dbgbuf);
  release(&dbglock);
}

// --- the device -------------------------------------------------------------

// A command. The answer replaces whatever was there.
//
// **The answer is read from offset zero.** That matters because writing to a
// descriptor advances its offset, so a program that writes and then reads on
// the same descriptor starts reading as many bytes into the answer as its
// command was long -- which is exactly the bug this comment exists to stop
// being written again. `cat` is unaffected because it opens the file fresh;
// cmd/schdbg seeks back to zero. See lseek() there.
int
dbg_write(struct inode *ip, uint64 off, char *src, int n)
{
  char line[128];
  int i;

  (void)ip; (void)off;

  if (n <= 0)
    return 0;
  if (n > (int)sizeof(line) - 1)
    n = (int)sizeof(line) - 1;
  for (i = 0; i < n; i++)
    line[i] = src[i];
  line[n] = 0;

  if (!dbginit) { initlock(&dbglock, "dbg"); dbginit = 1; }

  acquire(&dbglock);
  dbglen = fill(line);
  release(&dbglock);

  return n;                    // the whole line was consumed either way
}

int
dbg_read(struct inode *ip, uint64 off, char *dst, int n)
{
  int i, avail;

  (void)ip;

  if (!dbginit) { initlock(&dbglock, "dbg"); dbginit = 1; }

  acquire(&dbglock);
  if (dbglen == 0) {
    // Nothing asked yet. Answering with the help text rather than with EOF,
    // because `cat /dev/dbg` on its own is what someone does first and an
    // empty file teaches them nothing.
    struct sbuf sb = { dbgbuf, 0, DBGBUF };
    do_help(&sb);
    dbglen = sb.len;
  }
  if (off >= (uint64)dbglen) {
    release(&dbglock);
    return 0;                  // EOF
  }
  avail = dbglen - (int)off;
  if (n > avail)
    n = avail;
  for (i = 0; i < n; i++)
    dst[i] = dbgbuf[off + i];
  release(&dbglock);
  return n;
}