A static variable compiles with -O1 and above but not with -O0

[Adrian Meneses]

Hello, I'm trying to understand why a member static variable does not compile with the flag -O0 but it does with -O1 and above. A snippet code:
// This is the header file utest.h #ifndef UTEST #define UTEST namespace Builder { class Worker { public:    static const unsigned int SKILL = 10; }; } // end namespace Builder #endif
// main.cpp #include "utest.h" #include <iostream> #include <memory> int main() {    // I can compile this line just fine with -O0    std::cout << "SKILL is: " << Builder::Worker::SKILL << "\n";    // Compiling the following line with -O0 I get 'undefined reference to Builder::Worker::SKILL' but not with -O1 or -O2 for example... why?    const std::shared_ptr<unsigned int> myPtr = std::make_shared<unsigned int>(Builder::Worker::SKILL); }
I've been reading about One Definition Rule and it seems that the way SKILL is being used is the issue but I don't quite understand it. Why would the compiler complaint with -O0 but not with -O1? I know static variables should be declared in the cpp file always but I'm trying to understand the difference between -O0 and the other optimization flags. The command I used to compile was:
// For reference, I'm using compiler version g++ 12 and C++17 > g++ -Wall -Werror -O0 -o out main.cpp

[John Matthews]

Good question - I don't know, but it can be demonstrated using the Online C++ Compiler:
https://onlinegdb.com/g-czg9Hfu

Click on 'Run' and you get a compiler error. But click on 'Fork this', click on the settings cog button (top right), add -O1 to 'Extra Compiler Flags' and 'Run' again - it builds/runs fine.

The fact it can be persuaded to compile ok is probably a gcc-specific idiosyncrasy; if it didn't compile when it should then it would be a gcc bug.