linux-kernel - Re: [RFC/PATCH v2] ftrace: Reduce size of function graph entries

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Message-ID: <20160628215738.6e4815dd@gandalf.local.home>
Date:	Tue, 28 Jun 2016 21:57:38 -0400
From:	Steven Rostedt <rostedt@...dmis.org>
To:	Namhyung Kim <namhyung@...nel.org>
Cc:	Ingo Molnar <mingo@...nel.org>,
	LKML <linux-kernel@...r.kernel.org>, <linux-arch@...r.kernel.org>
Subject: Re: [RFC/PATCH v2] ftrace: Reduce size of function graph entries

On Wed, 29 Jun 2016 10:32:34 +0900
Namhyung Kim <namhyung@...nel.org> wrote:

> > is at 32-37
> > 
> > For a total of 38 bytes. I'm betting that without the packed, the 4
> > extra bytes will always be at the end.  
> 
> Woundn't it be 36 or 40 bytes? :)

Ug, don't know what I was counting then. I added the 64 bit version as
a after thought.

> 
> > 
> > If the compiler places it incorrectly without any attribute, it will
> > fail to read the long long if the arch requires 64 bits to be 8 bytes
> > aligned. The alignment is meaningless here. All we need is "packed" and
> > be done with it. It's only going to truncate the 4 bytes at the end of
> > the structure if that.  
> 
> I agree that in-struct alignment preserved without the 'aligned'
> attribute but I'm not sure whether it's guaranteed that the *start*
> address of the struct is still in proper alignment boundary.
> 
> IOW the struct ftrace_graph_ret should be placed at 8-byte boundary in order
> to keep alignment of struct members.  Is it guaranteed after applying
> 'packed'?
> 

My point is, a structure doesn't change size depending on where it is
located. Thus, if a structure contains a 8 byte field, that must be 8
bytes aligned due to architecture constraints, then it had better be
aligned that way everywhere. If it is not, then accessing the 8 byte
fields will cause issues.

The only thing that the "packed" changes, is it removes the last 4
padded bytes of the structure. It doesn't change anything else. Hence,
the alignment is just extra and unneeded.

-- Steve