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.