Does a function local static variable automatically incur a branch?

Question

For example:

int foo() {     static int i = 0;     return i++; } 

The variable i will only be initialized to 0 the first time foo is called. Does this automatically mean there's a hidden branch in there to keep the initialization from happening more than once? Or are there more clever tricks to avoid this?

Answer 1

Yes, it must incur a branch, and it must also incur at least an atomic operation for safe concurrent initialization. The Standard requires that they are initialized on function entry, in a concurrency-safe way.

The implementation can only dodge this requirement if it can prove that the difference between lazy init and some earlier initialization like before main() is entered is equivalent. For example, simple PODs initialized from constants, the compiler may choose to initialize it earlier like a file-scope global since it's non-observable and saving the lazy initialization code, but that's a non-observable optimization.