C++ scoped static initialization is not thread-safe, on purpose!

Raymond Chen


After this article was written,
the C++ standard has been revised.
Starting in C++11,
scoped static initialization is now thread-safe,
but it comes with a cost: Reentrancy now invokes undefined behavior.]

The rule for static variables at block scope
(as opposed to static variables with global scope)
is that they are initialized the first time execution
reaches their declaration.

Find the race condition:

int ComputeSomething()
  static int cachedResult = ComputeSomethingSlowly();
  return cachedResult;

The intent of this code is
to compute something expensive the first time the
function is called, and then cache the result to be
returned by future calls to the function.

A variation on this basic technique is

is advocated by this web site to avoid the “static initialization
order fiasco”
(Said fiasco is well-described on that page so I encourage you
to read it and understand it.)

The problem is that this code is not thread-safe. Statics
with local scope are internally converted by the compiler into
something like this:

int ComputeSomething()
  static bool cachedResult_computed = false;
  static int cachedResult;
  if (!cachedResult_computed) {
    cachedResult_computed = true;
    cachedResult = ComputeSomethingSlowly();
  return cachedResult;

Now the race condition is easier to see.

Suppose two threads both call this function for the first time.
The first thread gets as far as setting
cachedResult_computed = true,
and then gets pre-empted.
The second thread now sees that cachedResult_computed is true
and skips over the body of the “if” branch and returns
an uninitialized variable.

What you see here is not a compiler bug.
This behavior is required by the C++ standard.

You can write variations on this theme to create even worse

class Something { ... };
int ComputeSomething()
  static Something s;
  return s.ComputeIt();

This gets rewritten internally as
(this time, using pseudo-C++):

class Something { ... };
int ComputeSomething()
  static bool s_constructed = false;
  static uninitialized Something s;
  if (!s_constructed) {
    s_constructed = true;
    new(&s) Something; // construct it
  return s.ComputeIt();
// Destruct s at process termination
void DestructS()

Notice that there are multiple race conditions here.
As before, it’s possible for one thread to run ahead of the
other thread and use “s” before it has been constructed.

Even worse, it’s possible for the first thread to get
pre-empted immediately after testing s_constructed
but before setting it to “true”.
In this case, the object s gets double-constructed
and double-destructed.

That can’t be good.

But wait, that’s not all. Not look at what happens if you
have two runtime-initialized local statics:

class Something { ... };
int ComputeSomething()
  static Something s(0);
  static Something t(1);
  return s.ComputeIt() + t.ComputeIt();

This is converted by the compiler into the following

class Something { ... };
int ComputeSomething()
  static char constructed = 0;
  static uninitialized Something s;
  if (!(constructed & 1)) {
    constructed |= 1;
    new(&s) Something; // construct it
  static uninitialized Something t;
  if (!(constructed & 2)) {
    constructed |= 2;
    new(&t) Something; // construct it
  return s.ComputeIt() + t.ComputeIt();

To save space, the compiler placed the two
“x_constructed” variables into a bitfield.
Now there are multiple
read-modify-store operations on the variable

Now consider what happens if one thread
attempts to execute “constructed |= 1”
at the same time another thread attempts
to execute “constructed |= 2”.

On an x86, the statements likely assemble into

  or constructed, 1
  or constructed, 2

without any “lock” prefixes.
On multiprocessor machines, it is possible
for the two stores both to read the old value
and clobber each other with conflicting values.

On ia64 and alpha, this clobbering is much more
obvious since they do not have a single
read-modify-store instruction; the three
steps must be explicitly coded:

  ldl t1,0(a0)     ; load
  addl t1,1,t1     ; modify
  stl t1,1,0(a0)   ; store

If the thread gets pre-empted between the load
and the store, the value stored may no longer
agree with the value being overwritten.

So now consider the following insane sequence of execution:

  • Thread A tests “constructed” and finds it zero and prepares
    to set the value to 1, but it gets pre-empted.

  • Thread B enters the same function, sees “constructed” is zero
    and proceeds to construct both “s” and “t”, leaving
    “constructed” equal to 3.

  • Thread A resumes execution and completes its load-modify-store
    sequence, setting “constructed” to 1, then constructs “s”
    (a second time).

  • Thread A then proceeds to construct “t” as well (a second time)
    setting “constructed” (finally) to 3.

Now, you might think you can wrap the runtime initialization
in a critical section:

int ComputeSomething()
 static int cachedResult = ComputeSomethingSlowly();
 return cachedResult;

Because now you’ve placed the one-time initialization inside
a critical section and made it thread-safe.

But what if the second call comes from within the same thread?
(“We’ve traced the call; it’s coming from inside the thread!”)
This can happen if ComputeSomethingSlowly() itself calls
ComputeSomething(), perhaps indirectly.
Since that thread already owns the critical section, the code
enter it just fine and you once again
end up returning an uninitialized variable.

Conclusion: When you see runtime initialization of a local static
variable, be very concerned.

Raymond Chen
Raymond Chen

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