CODEWITHSUNDEEP
c++initialization
jbosch
jbosch

Reputation: 991

Are static data members safe as C++ default arguments?

Do I need to worry about the static initialization order fiasco when using a static data member as a default argument value? For example:

class Thing {

    static double const default_blarg;  // initialized in another file

    void run(double blarg=default_blarg);

};

I know default_blarg will be initialized at a mostly-unspecified point at link time, but I'm not sure when the default argument to run is initialized. If it's at some point that could be before default_blarg is initialized, what approach could I use to safely expose the default value as part of the class interface without repeating it? Would using constexpr for the static data member make it safe?

Note that I'm aware that this could yield some very confusing behavior if default_blarg wasn't const (that's why it is), and I'm not trying to use a non-static data member.

Upvotes: 9

Views: 660

Answers (2)

NathanOliver
NathanOliver

Reputation: 180595

You still have to be worried about the static initialization order fiasco. Lets say you have a.cpp and b.cpp. In a.cpp you have

double const Thing::default_blarg = 0;

Now in a.cpp any call to run after this point will have a initialized default and you are good to go. Unfortunately in b.cpp you have another static object that happens to create an instance of Thing and call run.

Now we do not know what will happen. If b.cpp runs first then default_blarg is not initialized and we have undefined behavior.

Now for you second part

but I'm not sure when the default argument to run is initialized

The default arguments of a function are evaluated each time the function is called. So as long as the function is called after default_blarg is initialized(and as noted above it may not) you will be okay.

Upvotes: 8

R Sahu
R Sahu

Reputation: 206587

From the C++11 Standard, Section 8.3.6/9:

Default arguments are evaluated each time the function is called.

As long as Thing::default_blarg is initialized before Thing::run is called, you should see predictable behavior.

Upvotes: 5

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