Are C++11 thread_local variables automatically static?

Question

Is there a difference between these two code segments:

void f() {     thread_local vector<int> V;     V.clear();     ... // use V as a temporary variable } 

and

void f() {     static thread_local vector<int> V;     V.clear();     ... // use V as a temporary variable } 

Backstory: originally I had a STATIC vector V (for holding some intermediate values, it gets cleared every time I enter the function) and a single-threaded program. I want to turn the program into a multithreading one, so somehow I have to get rid of this static modifier. My idea is to turn every static into thread_local and not worry about anything else? Can this approach backfire?

Answer 1

According to the C++ Standard

When thread_local is applied to a variable of block scope the storage-class-specifier static is implied if it does not appear explicitly

So it means that this definition

void f() {     thread_local vector<int> V;     V.clear();     ... // use V as a temporary variable } 

is equivalent to

void f() {     static thread_local vector<int> V;     V.clear();     ... // use V as a temporary variable } 

However, a static variable is not the same as a thread_local variable.

1 All variables declared with the thread_local keyword have thread storage duration. The storage for these entities shall last for the duration of the thread in which they are created. There is a distinct object or reference per thread, and use of the declared name refers to the entity associated with the current thread

To distinguish these variables the standard introduces a new term thread storage duration along with static storage duration.

Answer 2

Yes, "thread-local storage" is very similar to "global" (or "static storage"), only that instead of "duration of the entire program" you have "duration of the entire thread". So a block-local thread-local variable is initialized the first time control passes through its declaration, but separately within each thread, and it's destroyed when the thread ends.