Hi Stephen,
On 16 July 2018 at 16:51, Stephen Warren swarren@wwwdotorg.org wrote:
From: Stephen Warren swarren@nvidia.com
The U-Boot Makefile can invoke binman multiple times in parallel. This is problematic because binman uses a static hard-coded temporary file name. If two instances of binman use that filename at the same time, one writing one reading, they may silently read the wrong content or actively detect missing signatures and error out the build process. Fix this by using a PID-specific filename instead.
Signed-off-by: Stephen Warren swarren@nvidia.com
tools/binman/control.py | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-)
diff --git a/tools/binman/control.py b/tools/binman/control.py index a40b300fdacb..515999278949 100644 --- a/tools/binman/control.py +++ b/tools/binman/control.py @@ -121,7 +121,7 @@ def Binman(options, args): # output into a file in our output directly. Then scan it for use # in binman. dtb_fname = fdt_util.EnsureCompiled(dtb_fname)
fname = tools.GetOutputFilename('u-boot-out.dtb')
fname = tools.GetOutputFilename('u-boot-out.dtb') + str(os.getpid()) with open(dtb_fname) as infd: with open(fname, 'wb') as outfd: outfd.write(infd.read())-- 2.18.0
But the output directory is itself (normally) a temporary dir. That determines the directly which GetOutputFilename() uses. So I am not sure how this can happen in practice?
Regards, Simon