c++ singleton with static variable allocated with new [duplicate]

Question

recently I saw a piece of code as follows :

namespace {
    mutex* get_server_factory_lock() {
        static mutex server_factory_lock;
        return &server_factory_lock;
    }

    typedef std::unordered_map<string, ServerFactory*> ServerFactories;
    ServerFactories* server_factories() {
        static ServerFactories* factories = new ServerFactories; 
        // is variable factories assigned every time this function called ? 
        return factories;
    }
}  // namespace

/* static */
void ServerFactory::Register(const string& server_type,
                         ServerFactory* factory) {
    mutex_lock l(*get_server_factory_lock());
    if (!server_factories()->insert({server_type, factory}).second) {
        LOG(ERROR) << "Two server factories are being registered under "
            << server_type;
    }
}

It seems that function server_factories() is similar to the singleton.
My question is : to the best of my knowledge, factories is a static variable, and every time function server_factories() called, this static variable will be assigned a new value. But the result is not, every time server_factories() is called, it returns the same pointer. Why?

PS : with c++11 enabled when compiling.

Duplicated with What is the lifetime of a static variable in a C++ function?

In both functions, static variable is only initialized once. — Jarod42
– Jarod42, Commented Jun 21, 2016 at 14:21
@Jarod42 Ok, thanks. But I still wonder why static ServerFactories* factories = new ServerFactories; is considered as initialization, not an assignment when server_factories() called in the second time. — pgplus1628
– pgplus1628, Commented Jun 21, 2016 at 15:29
Look at whats-the-difference-between-assignment-operator-and-copy-constructor — Jarod42
– Jarod42, Commented Jun 21, 2016 at 15:43
@Jarod42 that link didn't help. I think it's an assignment operation, not copy operation. If so, the factories variable should be changed every time server_factories() is called. — pgplus1628
– pgplus1628, Commented Jun 22, 2016 at 2:22

lorro · Accepted Answer · 2016-06-23 18:43:56Z

2

It's a static, so it only gets initialized once, when you first enter the function. Later calls to the same function use the previous variable. That said, I fail to see why it's a pointer and not a simple function static variable (with automatic storage duration) that we could take the address of...

edited Jun 23, 2016 at 18:43

answered Jun 21, 2016 at 14:21

lorro

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5 Comments

Dark Falcon Over a year ago

"That said, I fail to see why it's a pointer and not a simple static that we could take the address of" << Because perhaps ServerFactories() uses other global variables. If it were also a global, initialization order across translation units would be undefined. By making it a function-local static, initialization order becomes well-defined. Whether this is needed in his case is unknown, but this is a frequent reason for this pattern.

pgplus1628 Over a year ago

@DarkFalcon thanks.

Jarod42 Over a year ago

@DarkFalcon: in get_server_factory_lock(), there is no new, and you can still have well defined behavior. There is no points to do allocation.

lorro Over a year ago

@DarkFalcon: by simple static I mean static within the function, but not pointer (sorry if my wording was confusing), like this: ServerFactories* server_factories() { static ServerFactories factories; return &factories; } Editing the answer accordingly.

Dark Falcon Over a year ago

@lorro, fair enough. Might want to clarify that. The correct terminology is (I think) "function static variable with automatic storage duration"

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