How does magic statics ensure that the right side is only executed once?
If I have
atomic<int> cnt=0;
int get_int() noexcept
{
cnt++;
return rand();
}
and then:
void func()
{
static const auto value = get_int();
}
I know that value
there will be no race conditions on initialization , but I don't know if it get_int()
will be called once or if
it will cnt
be 1 in my example (not 2, 3, 4 or 5).
Suppose there are several threads coming in func()
, but get_int
only has one callsite in func()
.
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C ++ 11 guarantees there won't be any race conditions N3797 - §6.7 / 4:
An implementation is allowed to perform early initialization of other block scope variables with static or thread storage duration under the same conditions as the implementation is allowed to statically initialize a variable with static or thread storage duration in the namespace (3.6.2). Otherwise, such a variable initialized first time control passes through its declaration; such a variable is considered initialized by the completion of its initialization. If initialization is completed by throwing an exception, the initialization is not complete, so it will be checked again the next time the control enters the declaration. If control enters the declaration concurrently when the variable is initialized, the concurrent execution must wait for the initialization to complete.92 If control re-enters the declaration recursively while the variable is initialized, the behavior is undefined. [Example:
int foo(int i) { static int s = foo(2*i); // recursive call - undefined return i+1; }
- end of example]
It is not re-entry, but thread safe. Make sure there are no other parts of the code that will be called get_int()
before func()
anyway.
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