On Fri, 02 Aug 2013 22:33:24 +0200 Andreas Färber <afaer...@suse.de> wrote:
> Am 23.07.2013 18:22, schrieb Igor Mammedov: > > Signed-off-by: Igor Mammedov <imamm...@redhat.com> > > --- > > vl.c | 7 +------ > > 1 files changed, 1 insertions(+), 6 deletions(-) > > > > diff --git a/vl.c b/vl.c > > index 8190504..bf0c658 100644 > > --- a/vl.c > > +++ b/vl.c > > @@ -2947,7 +2947,7 @@ int main(int argc, char **argv, char **envp) > > module_call_init(MODULE_INIT_MACHINE); > > machine = find_default_machine(); > > cpu_model = NULL; > > - ram_size = 0; > > + ram_size = DEFAULT_RAM_SIZE * 1024 * 1024; > > snapshot = 0; > > cyls = heads = secs = 0; > > translation = BIOS_ATA_TRANSLATION_AUTO; > > @@ -4064,11 +4064,6 @@ int main(int argc, char **argv, char **envp) > > exit(1); > > } > > > > - /* init the memory */ > > - if (ram_size == 0) { > > - ram_size = DEFAULT_RAM_SIZE * 1024 * 1024; > > - } > > - > > if (qemu_opts_foreach(qemu_find_opts("device"), device_help_func, > > NULL, 0) > > != 0) { > > exit(0); > > Commit message doesn't give any explanation why? it was intended as cleanup > > What happens with -m 0? My guess is the old code translates that to the > default size, where by intializing the default earlier it would stay. patch is broken in this aspect. It aborts on start up with -m 0 The question is if -m 0 is correct value, perhaps QEMU should exit with error message in this case, instead of silent fallback to default? > > Andreas >