gcc -g调试标志是否会影响程序执行？

Question

I've just been testing a program I'm working, and I see that it's executing 3μs faster (a statistically significant change) when I compile it with -g. This makes no sense to me - I thought that the -g flag wasn't supposed to affect the program execution, and that even if it did it would make it run slower, not faster.

Can anyone tell me why this is happening? And whether it changes the programs execution flow? I am not compiling with -O because I need it to execute exactly as written, but if -g can somehow make it run faster with changing the instruction order I should obviously be using that.

So I need to know exactly what changes the -g flag makes to the program.

Edit: The more tests I run, the bigger the t-value gets (= the more statistically significant the difference becomes). This is definitely not measurement error - something is going on.

Answer 1

As others have said, debugging symbols will not change the control flow of your code unless that there is an (unlikely) bug in the compiler.

It changes execution, though, because the executable becomes bigger, and the executed code is spread more widely on more pages. You can expect more cache misses and IO signals. On a multi-tasking environment (and even a Linux/busybox system is such a thing) this can result is slightly different scheduling behavior.

On the other hand, measuring such tiny time differences as you describe them is an art in its own rights. You are probably in an Heisenberg setting, where your measurements influence execution times. Your measurements may show statistically significant deviation, but I would be extremely careful in interpreting them as saying such and such option makes faster code.

Answer 2

The -g flag makes 0 changes to the actual generated code. What it does is add debug sections to the executable. Those sections are not loaded at runtime, but debuggers can load them. As the executable now is a bit different it's larger - you might try to measure the no. of page faults going on with one version vs another., there will be changes in how the executable is stored on disk, but no code changes.

If you want to see the assembly, run objdump -d on your binary and compare

I do question the validity of the 3us increase though, reliably measuring 3us, at least on a general purpose OS is a hard task - I hope you have run your program a few thousand times (likely a few hundred thousand times rather) to come up with that number to try to eliminate all the random things affecting such a measurement.

Answer 3

当我在其中一个子程序上使用-debug和-g标志时，我的代码得到了不同的答案，所以虽然我不知道为什么，是的，它肯定会影响程序的执行。