The kernel is compiled with -Os for code size last I checked (which was long ago) so that might explain the difference.
It's been about ten years since I did any kernel work, but last I checked Linux doesn't use most of the fancy features of the CPU to avoid trashing more registers than necessary. If you know you don't do any floating point operations, for example, you know you don't need to save/restore the floating point registers on interrupts. The same might go for things like SSE.
Those instructions tend to be larger, so it may partly explain the difference in average instruction size. The same explanation would apply to why the Radeon driver has a higher average.
I wonder if that flag does anything at the instruction encoding level
Interesting, would you mind sharing the script or give a short overview how it works?
Here it is (warning: Perl): http://pastebin.com/MFmDzY0g
It calls "objdump -d" on the binary, which is binutil's disassembler.
objdump prints each instruction out in the following format:
5dc: 67 80 7d 00 00 cmpb $0x0,0x0(%ebp)
The script uses regexs to cut off the front and back, then counts the number of bytes (in hex format) in the second part.
Then it calculates the average for all instructions.
Might I suggest http://pastebin.com/2PHXi78A instead? It handles a few bugs from the original, and cleans the nested loops.
I was curious about this claim so I wrote a quick and dirty perl script to test it out:
Seems spot on, except for the kernel. I'm not sure why that has shorter instructions on average.EDIT: I ran it on git too; average was 3.96747336803081. It seems Linus' x86 wizardry does not extend to userspace in this case.