1 files changed, 59 insertions, 69 deletions

diff --git a/sysdeps/unix/sysv/linux/powerpc/get_clockfreq.c b/sysdeps/unix/sysv/linux/powerpc/get_clockfreq.c
index 9b8165d203..fe8c5a15b3 100644
--- a/sysdeps/unix/sysv/linux/powerpc/get_clockfreq.c
+++ b/sysdeps/unix/sysv/linux/powerpc/get_clockfreq.c
@@ -24,95 +24,85 @@
 #include <libc-internal.h>
 #include <sysdep.h>
 #include <bits/libc-vdso.h>
+#include <not-cancel.h>
 
 hp_timing_t
 __get_clockfreq (void)
 {
+  hp_timing_t result = 0L;
+
+#ifdef SHARED
+  /* The vDSO does not return an error (it clear cr0.so on returning).  */
+  INTERNAL_SYSCALL_DECL (err);
+  result =
+    INTERNAL_VSYSCALL_NO_SYSCALL_FALLBACK (get_tbfreq, err, uint64_t, 0);
+#else
   /* We read the information from the /proc filesystem.  /proc/cpuinfo
      contains at least one line like:
      timebase        : 33333333
      We search for this line and convert the number into an integer.  */
-  static hp_timing_t timebase_freq;
-  hp_timing_t result = 0L;
+  int fd = __open_nocancel ("/proc/cpuinfo", O_RDONLY);
+  if (__glibc_likely (fd != -1))
+    return result;
 
-  /* If this function was called before, we know the result.  */
-  if (timebase_freq != 0)
-    return timebase_freq;
+  /* The timebase will be in the 1st 1024 bytes for systems with up
+     to 8 processors.  If the first read returns less then 1024
+     bytes read,  we have the whole cpuinfo and can start the scan.
+     Otherwise we will have to read more to insure we have the
+     timebase value in the scan.  */
+  char buf[1024];
+  ssize_t n;
 
-  /* If we can use the vDSO to obtain the timebase even better.  */
-#ifdef SHARED
-  INTERNAL_SYSCALL_DECL (err);
-  timebase_freq =
-    INTERNAL_VSYSCALL_NO_SYSCALL_FALLBACK (get_tbfreq, err, uint64_t, 0);
-  if (INTERNAL_SYSCALL_ERROR_P (timebase_freq, err)
-      && INTERNAL_SYSCALL_ERRNO (timebase_freq, err) == ENOSYS)
-#endif
+  n = __read_nocancel (fd, buf, sizeof (buf));
+  if (n == sizeof (buf))
     {
-      int fd = __open ("/proc/cpuinfo", O_RDONLY);
+      /* We are here because the 1st read returned exactly sizeof
+         (buf) bytes.  This implies that we are not at EOF and may
+         not have read the timebase value yet.  So we need to read
+         more bytes until we know we have EOF.  We copy the lower
+         half of buf to the upper half and read sizeof (buf)/2
+         bytes into the lower half of buf and repeat until we
+         reach EOF.  We can assume that the timebase will be in
+         the last 512 bytes of cpuinfo, so two 512 byte half_bufs
+         will be sufficient to contain the timebase and will
+         handle the case where the timebase spans the half_buf
+         boundry.  */
+      const ssize_t half_buf = sizeof (buf) / 2;
+      while (n >= half_buf)
+	{
+	  memcpy (buf, buf + half_buf, half_buf);
+	  n = __read_nocancel (fd, buf + half_buf, half_buf);
+	}
+      if (n >= 0)
+	n += half_buf;
+    }
+  __close_nocancel (fd);
 
-      if (__glibc_likely (fd != -1))
+  if (__glibc_likely (n > 0))
+    {
+      char *mhz = memmem (buf, n, "timebase", 7);
+
+      if (__glibc_likely (mhz != NULL))
 	{
-	  /* The timebase will be in the 1st 1024 bytes for systems with up
-	     to 8 processors.  If the first read returns less then 1024
-	     bytes read,  we have the whole cpuinfo and can start the scan.
-	     Otherwise we will have to read more to insure we have the
-	     timebase value in the scan.  */
-	  char buf[1024];
-	  ssize_t n;
+	  char *endp = buf + n;
 
-	  n = __read (fd, buf, sizeof (buf));
-	  if (n == sizeof (buf))
-	    {
-	      /* We are here because the 1st read returned exactly sizeof
-	         (buf) bytes.  This implies that we are not at EOF and may
-	         not have read the timebase value yet.  So we need to read
-	         more bytes until we know we have EOF.  We copy the lower
-	         half of buf to the upper half and read sizeof (buf)/2
-	         bytes into the lower half of buf and repeat until we
-	         reach EOF.  We can assume that the timebase will be in
-	         the last 512 bytes of cpuinfo, so two 512 byte half_bufs
-	         will be sufficient to contain the timebase and will
-	         handle the case where the timebase spans the half_buf
-	         boundry.  */
-	      const ssize_t half_buf = sizeof (buf) / 2;
-	      while (n >= half_buf)
-		{
-		  memcpy (buf, buf + half_buf, half_buf);
-		  n = __read (fd, buf + half_buf, half_buf);
-		}
-	      if (n >= 0)
-		n += half_buf;
-	    }
+	  /* Search for the beginning of the string.  */
+	  while (mhz < endp && (*mhz < '0' || *mhz > '9') && *mhz != '\n')
+	    ++mhz;
 
-	  if (__builtin_expect (n, 1) > 0)
+	  while (mhz < endp && *mhz != '\n')
 	    {
-	      char *mhz = memmem (buf, n, "timebase", 7);
-
-	      if (__glibc_likely (mhz != NULL))
+	      if (*mhz >= '0' && *mhz <= '9')
 		{
-		  char *endp = buf + n;
-
-		  /* Search for the beginning of the string.  */
-		  while (mhz < endp && (*mhz < '0' || *mhz > '9')
-			 && *mhz != '\n')
-		    ++mhz;
-
-		  while (mhz < endp && *mhz != '\n')
-		    {
-		      if (*mhz >= '0' && *mhz <= '9')
-			{
-			  result *= 10;
-			  result += *mhz - '0';
-			}
-
-		      ++mhz;
-		    }
+		  result *= 10;
+		  result += *mhz - '0';
 		}
-	      timebase_freq = result;
+
+	      ++mhz;
 	    }
-	  __close (fd);
 	}
     }
+#endif
 
-  return timebase_freq;
+  return result;
 }