From aded6539d983280212e08d09f14157b1cb4d58cc Mon Sep 17 00:00:00 2001 From: Juan Quintela Date: Tue, 11 Feb 2014 22:56:00 +0100 Subject: [PATCH] qemu_file: use fwrite() correctly fwrite() returns the number of items written. But when there is one error, it can return a short write. In the particular bug that I was tracking, I did a migration to a read-only filesystem. And it was able to finish the migration correctly. fwrite() never returned a negative error code, nor zero, always 4096. (migration writes chunks of about 14000 bytes). And it was able to "complete" the migration with success (yes, reading the file was a bit more difficult). To add insult to injury, if your amount of memory was big enough (12GB on my case), it overwrote some important structure, and from them, malloc failed. This check makes the problem go away. Signed-off-by: Juan Quintela Reviewed-by: Eric Blake Reviewed-by: Markus Armbruster Signed-off-by: Juan Quintela --- qemu-file.c | 9 ++++++++- 1 file changed, 8 insertions(+), 1 deletion(-) diff --git a/qemu-file.c b/qemu-file.c index 9473b674ba..f074af15c3 100644 --- a/qemu-file.c +++ b/qemu-file.c @@ -100,7 +100,14 @@ static int stdio_put_buffer(void *opaque, const uint8_t *buf, int64_t pos, int size) { QEMUFileStdio *s = opaque; - return fwrite(buf, 1, size, s->stdio_file); + int res; + + res = fwrite(buf, 1, size, s->stdio_file); + + if (res != size) { + return -EIO; /* fake errno value */ + } + return res; } static int stdio_get_buffer(void *opaque, uint8_t *buf, int64_t pos, int size) -- GitLab