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When I want to do debugging of C or C++ programs, I've been taught to use -O0 to turn optimization OFF, and -ggdb to insert symbols into the executable which are optimized for using the GNU gdb debugger, which I use (or, you can use -glldb for LLVM/clang's lldb debugger, or just -g for general debugging symbols, but that won't be as good as -ggdb apparently...). However, I recently stumbled upon someone saying to use -Og (instead of -O0), and it caught me off-guard. Sure enough though, it's in man gcc!:

-Og Optimize debugging experience. -Og enables optimizations that do not interfere with debugging. It should be the optimization level of choice for the standard edit-compile-debug cycle, offering a reasonable level of optimization while maintaining fast compilation and a good debugging experience.

So, what's the difference? Here's the -O0 description from man gcc:

-O0 Reduce compilation time and make debugging produce the expected results. This is the default.

man gcc clearly says -Og "should be the optimization level of choice for the standard edit-compile-debug cycle", though.

This makes it sound like -O0 is truly "no optimizations", whereas -Og is "some optimizations on, but only those which don't interfere with debugging." Is this correct? So, which should I use, and why?

Related:

  1. related, but NOT a duplicate! (read it closely, it's not at all a duplicate): What is the difference between -O0 ,-O1 and -g
  2. my answer on debugging --copt= settings to use with Bazel: gdb: No symbol "i" in current context
Gabriel Staples
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    [The manual](https://gcc.gnu.org/onlinedocs/gcc/Optimize-Options.html#Optimize-Options) has more details: `Like -O0, -Og completely disables a number of optimization passes so that individual options controlling them have no effect. Otherwise -Og enables all -O1 optimization flags except for those that may interfere with debugging:`. Then it lists the exact flags disabled. – kaylum Aug 13 '20 at 00:04
  • Ah, that is useful! That's more information than is in `man gcc`! – Gabriel Staples Aug 13 '20 at 00:09
  • If you read the excerpts in your question carefully, I think the answer is there quite explicitly. `-Og` optimizes the debugging experience, whereas `-O0` minimizes the compilation speed. You want to use `-Og` if you want to enjoy debugging, `-O0` if you want to quickly find out if the damn thing even compiles. – Antti Haapala -- Слава Україні Aug 13 '20 at 04:23
  • @AnttiHaapala, true, but it's nice to have confirmation from other developers, especially when I'm so surprised by the new information (ie: that `-Og` even exists). – Gabriel Staples Aug 13 '20 at 19:01

1 Answers1

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@kaylum just provided some great insight in their comment under my question! And the key part I really care about the most is this:

[-Og] is a better choice than -O0 for producing debuggable code because some compiler passes that collect debug information are disabled at -O0.

https://gcc.gnu.org/onlinedocs/gcc/Optimize-Options.html#Optimize-Options

So, from now on I'm using -Og (NOT -O0) in addition to -ggdb.

UDPATE 13 Aug. 2020:

Heck with this! Nevermind. I'm sticking with -O0.

With -Og I get <optimized out> and Can't take address of "var" which isn't an lvalue. errors all over the place! I can't print my variables or examine their memory anymore! Ex:

(gdb) print &angle
Can't take address of "angle" which isn't an lvalue.
(gdb) print angle_fixed_p
$6 = <optimized out>

With -O0, however, everything works fine!

(gdb) print angle
$7 = -1.34869879e+20
(gdb) print &angle
$8 = (float *) 0x7ffffffefbbc
(gdb) x angle
0x8000000000000000:     Cannot access memory at address 0x8000000000000000
(gdb) x &angle
0x7ffffffefbbc: 0xe0e9f642

So, back to using -O0 instead of -Og it is!

Related:

  1. [they also recommend -O0, and I concur] What does <value optimized out> mean in gdb?
Gabriel Staples
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