Patchwork [v3,08/19] qemu-img: always probe the input image for allocated sectors

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Submitter Paolo Bonzini
Date July 25, 2013, 2:23 p.m.
Message ID <1374762197-7261-9-git-send-email-pbonzini@redhat.com>
Download mbox | patch
Permalink /patch/261720/
State New
Headers show

Comments

Paolo Bonzini - July 25, 2013, 2:23 p.m.
qemu-img convert is assuming "that sectors which are unallocated in the input
image are present in both the output's and input's base images", but it is
only doing this if the output image is zero initialized.  And checking if
the output image is zero initialized does not make much sense if the
output image is copy-on-write.  Always do the test.

Reviewed-by: Eric Blake <eblake@redhat.com>
Signed-off-by: Paolo Bonzini <pbonzini@redhat.com>
---
 qemu-img.c | 26 ++++++++++++--------------
 1 file changed, 12 insertions(+), 14 deletions(-)

Patch

diff --git a/qemu-img.c b/qemu-img.c
index a4957eb..3faffee 100644
--- a/qemu-img.c
+++ b/qemu-img.c
@@ -1476,21 +1476,19 @@  static int img_convert(int argc, char **argv)
                 n = bs_offset + bs_sectors - sector_num;
             }
 
-            if (has_zero_init) {
-                /* If the output image is being created as a copy on write image,
-                   assume that sectors which are unallocated in the input image
-                   are present in both the output's and input's base images (no
-                   need to copy them). */
-                if (out_baseimg) {
-                    if (!bdrv_is_allocated(bs[bs_i], sector_num - bs_offset,
-                                           n, &n1)) {
-                        sector_num += n1;
-                        continue;
-                    }
-                    /* The next 'n1' sectors are allocated in the input image. Copy
-                       only those as they may be followed by unallocated sectors. */
-                    n = n1;
+            /* If the output image is being created as a copy on write image,
+               assume that sectors which are unallocated in the input image
+               are present in both the output's and input's base images (no
+               need to copy them). */
+            if (out_baseimg) {
+                if (!bdrv_is_allocated(bs[bs_i], sector_num - bs_offset,
+                                       n, &n1)) {
+                    sector_num += n1;
+                    continue;
                 }
+                /* The next 'n1' sectors are allocated in the input image. Copy
+                   only those as they may be followed by unallocated sectors. */
+                n = n1;
             } else {
                 n1 = n;
             }