On static variables, static storage, static initialization, constant initialization, constinit, constexpr

In a recent discussion over Twitter, it was pointed out that optimizers failed to eliminate a function-scope static variable with no uses. This article explores why the optimizer struggles with such code patterns, how static variables are stored and initialized, and also how certain C++ keywords can help the optimizer do its job.

Disclaimer: Godbolt links will be used but, instead of inspecting x86 assembly, the target will be LLVM's Intermediate Representation (IR). It is almost always simpler to use IR instead of assembly:

Most operations and types are spelled out explicitly.

The data section is easier to visualize.

The compiler transformations we're interested in are, more often than not, architecture-independent and happen before the compiler generates assembly. Therefore we can keep a higher level of abstraction (IR) that is easier to reason about.

We get to see what the compiler does to obey the C++ standard much earlier: all the rules must be captured in the translation from C++ to IR and, from there on, all optimizations are game - the standard doesn't exist anymore.

Everything necessary about IR will be explained, but if you want to learn more, I presented a tutorial during the EuroLLVM developers conference in 2019.

The offending code.

The original tweet used this example:

int static_version () { static const std :: string unused_variable{ "static" }; return 42 ; } int local_version () { const std :: string unused_variable{ "local" }; return 42 ; }

The author mentions how, in the static case, unused_variable doesn't get optimized away but, in the local variable case, the optimizer does a much better job.

Because std::string has a complicated constructor, I'll rewrite this code using the simplest class possible:

struct State { State() {} }; int get_value () { static State optimize_me; return 42 ; }

It's reasonable to expect that State optimize_me will be optimized away: it is a function-scope static variable with no uses. Unfortunately, both GCC and Clang fail to do so.

To understand what's going here, let's look at the IR produced by Clang without optimizations1.

First, a type struct.State is defined:

%struct.State = type { i8 }

Our C++ struct has no data members, and yet its equivalent in IR contains an 8-bit integer ( i8 ). What is sizeof(State) ? Having seen the IR, the answer is easy to guess: 1 byte2.

Then, a global variable of that type is defined and initialized:

@_ZZ10get_value2vE11optimize_me = internal global %struct.State zeroinitializer

Note that this variable is initialized with the zeroinitializer keyword. That means its memory region will be set to zero before the program starts. But... We haven't initialized our C++ variable at all! We'll talk more about this later.

There is another global variable in our module, with a very similar name:

@_ZGVZ10get_value2vE11optimize_me = internal global i8 0

Note again how this variable is zero initialized, this time by writing i8 0 (this is equivalent to zeroinitializer ).

This is very mysterious: an 8-bit integer that we never wrote in the original C++ code. Let's look at the body of our get_value function to find out more:

define i32 @_Z10get_valuev() #0 { %1 = load i8, i8* @_ZGVZ10get_value2vE11optimize_me %2 = icmp eq i8 %1, 0 br i1 %2, label %3, label %4

The first line loads the mysterious variable and the second line compares it to 0. If the value is zero, the code branches to this block:

3: call void @_ZN5StateC2Ev(%struct.State* @_ZZ10get_value2vE11optimize_me) store i8 1, i8* @_ZGVZ10get_value2vE11optimize_me br label %4

A function call with our static variable as its first argument -- this is a call to State 's constructor! Right after that, we update the value of the mysterious variable by storing 1 to it.

Afterwards, or if the original comparison to zero failed, code execution proceeds to return 42 :

4: ret i32 42 }

A visual representation can be found in the control flow graph for this function:

This example illustrates the code generated in order to initialize function-scope static variables. The compiler must guarantee that the constructor is called exactly once, during the first time execution passes through the static variable declaration. This is accomplished with the pattern:

Global counter initialized to 0

If counter is zero: Call constructor. Set counter to 1.



Optimized code

Is the optimizer able to remove all of the code in that function? Take a look:

@_ZGVZ9get_valuevE11optimize_me = internal unnamed_addr global i1 false define i32 @_Z9get_valuev() { %1 = load i1, i1* @_ZGVZ9get_valuevE11optimize_me br i1 %1, label %3, label %2 2: store i1 true, i1* @_ZGVZ9get_valuevE11optimize_me br label %3 3: ret i32 42 }

Note that the optimizer:

Deleted the static variable. Deleted the call to its constructor. Transformed the global counter into a boolean (from i8 to i1 ). Failed to optimize away this global boolean.

Point #4 is a hard problem because the counter will make the first call to get_value take a different code path from subsequent calls. Furthermore, the two paths have distinct behaviors: one writes to a global variable, the other doesn't. To delete the counter:

The optimizer needs to prove that the change in the counter's value isn't meaningful to the program. But it is meaningful because it affects control flow inside get_value . So the optimizer needs to prove that control flow inside get_value isn't meaningful to the program. But it is meaningful because control flow affects the counter's value.

... And now we're stuck in a loop! It's not an unsolvable problem, but it illustrates challenges the optimizer can't overcome right now3.

This situation gets worse if the constructor call isn't as simple as an empty function. Our motivating example, std::string , definitely doesn't have a simple constructor.

Can we do better ?

We can. We can help the compiler by expressing our intent more appropriately. But first, we need to understand how static variables are initialized.

Static storage

Static variables have what is known as static storage. From cppreference:

The storage for the object is allocated when the program begins and deallocated when the program ends. Only one instance of the object exists.

In practice, we see the storage for static variables in the data segment of the program, in other words, the storage is available when the program is loaded. Moreover, the initial contents of that memory region are also specified in the data segment and available when the program is loaded; but what exactly are those contents and can we influence them?

Zero or Constant initialization

Let's look at what cppreference tells us (emphasis mine):

Variables declared at block scope with the specifier static[...] are initialized the first time control passes through their declaration( unless their initialization is zero - or constant - initialization, which can be performed before the block is first entered) . [static local variables]

Intuitively, zero-initialization is what it sounds like: when the program is loaded, that region of memory gets zero initialized. Typically, if other initialization is necessary, like running constructors or evaluating constructor arguments, it will happen at runtime. Not very exciting.

Constant-initialization, when possible, happens instead of zero-initialization. The details are complicated, but it essentially boils down to whether you have a constant expression initializing the static variable. Cppreference uses the following notation to explain this idea:

static T object = constexpr;

The best part is that constant-initialization will typically remove the need for runtime initialization. Let's look at what this looks like in IR.

Constant initialization to the rescue!

Let's make our example slightly more complicated, disable all optimizations, but have a constexpr constructor:

struct State { constexpr State(char c1, char c2, char c3) : value1{c1}, value2{c2}, value3{c3} {} char value1; char value2; char value3; }; int get_value() { static State optimize_me(1, 2, 3); return 42; }

constexpr functions are a mechanism through which programmers express their desire to have the function evaluated at compile time if the function is called with compile time constant arguments.4

Because our static variable is now initialized with a constant expression, the IR for this function now becomes much simpler:

@_ZZ9get_valuevE11optimize_me = internal global %struct.State { i8 1, i8 2, i8 3 } define i32 @_Z10get_valuev() { ret i32 42 }

To emphasize, this happens with no optimizations, this is a built in mechanism of the language, not a compiler transformation. See for yourself!

What happened here? Constant initialization took place, because we have a constant expression (the constructor is constexpr and it is called with constant arguments) initializing the optimize_me variable.

In the non- constexpr version, the IR global variable corresponding to the C++ static variable was initialized by zeroinitializer , and inside the get_value function we had a constructor call wrapped by some boilerplate to ensure the variable was initialized exactly once. In other words, zero initialization + runtime initialization took place.

In the constexpr version, all the boilerplate is gone because constant initialization happened instead of zero-initialization + runtime initialization. This is the core idea of this post: if you enable constant initialization, unnecessary code disappears.

The generated assembly contains the already-initialized variable in the data segment of the program:

.data get_value()::optimize_me: .byte 1 # 0x1 .byte 2 # 0x2 .byte 3 # 0x3 .size get_value()::optimize_me, 3

With optimizations enabled, the static variable will be completely removed.

Don't let slow code compile.

C++20 adds a new keyword constinit to ensure a variable only has constant initialization, otherwise the program is ill-formed. For example, the following code does not compile (note the absence of a constexpr constructor):

struct State { State( char c1, char c2, char c3) : value1{c1}, value2{c2}, value3{c3} {} char value1; char value2; char value3; }; int get_value () { constinit static State optimize_me ( 1 , 2 , 3 ) ; return 42 ; }

This is desirable because it prevents inefficient code from compiling. If we make State 's constructor constexpr , the program is now legal and uses efficient constant initialization. Godbolt link

But we can't constexpr all the things

The original example dealt with a std::string static variable, which may perform dynamic memory allocation - which is not allowed in constexpr contexts. This is lifted in C++20 and most methods of std::string are made constexpr thanks to [Louis Dionne's paper]. No compilers implement this at the time of writing, but you can check GCC's progress and Clang's progress on their websites.

Edit (2020-03-21): As Jason Turner pointed out on Twitter, constexpr dynamic allocation, while allowed in C++ 20, still needs to be freed in the same constexpr context that allocated it. This implies that big constexpr strings are not going to be allowed.

Conclusion

Without entering the discussion of when/if static variables should be used, it's important to be aware of the price that is paid for their correct initialization. In most cases, the programmer can completely avoid this price by using constant initialization (usually in the form of constexpr constructors and functions).

Furthermore, by expressing their intent properly to the compiler, it's possible to ensure a compilation error when code changes trigger inefficient initialization; this is accomplished by marking the static variable as constinit . A lot of new features in the C++ language are driven by the desire to allow programmers to communicate intent to the compiler (and to other programmers).

I also hope to have shown that using the LLVM IR makes it simpler to explore architecture-agnostic missed optimizations. In the case explored here, there is no reason why a static variable should be optimized away when targeting x86, but not when targeting ARM, for instance.

1 We're compiling the code without support for thread-safe static initialization to keep things simple. However, most of our conclusion still hold if we enable thread safe statics↩

2If you're curious why, the creator of C++ answers it in his website.↩

3 Other challenges are possible. For example, if this function gets inlined elsewhere, we will have multiple functions accessing the same global variable and the compiler will struggle reasoning about this. Note also that we don't have to consider other translation units because static variables have internal linkage, that is, they can only be accessed from the translation unit in which it is defined; this is represented by the internal keyword in IR. ↩