In C++, what do braces on the left-hand side of a variable declaration mean?

Question

The code at this GitHub file uses a C++ variable "declaration" syntax I'm not familiar with:

std::unique_ptr<CRecentFileList> {m_pRecentFileList} = std::make_unique<CRecentFileList>(...

(m_pRecentFileList is declarared in a superclass.)

What does it mean when you wrap a variable declaration in braces? (not an initializer list)

---

I extracted a minimal test case which compiles:

class foo {
    int* p;
    void f(){
        std::unique_ptr<int> {p} = std::make_unique<int>(1);
    }
};

Changing int* p to std::unique_ptr<int> p creates a compilation error due to unique_ptr(const unique_ptr&) = delete;

This makes me think braced declaration assigns to a outer-scope variable with the same name. I tried creating a test program, but it fails to compile:

int main(){
    int x;
    int {x} = 1;
}

error: using temporary as lvalue [-fpermissive]

Answer 1

It's not a declaration. It's an assignment to a temporary.

In std::unique_ptr<int> {p} = std::make_unique<int>(1);, std::unique_ptr<int> {p} creates a unique_ptr temporary that takes ownership of the object p points to, then std::make_unique<int>(1) is assigned to that temporary, which causes the object p points to to be deleted and the temporary to take ownership of the int created by the make_unique; finally, at the ;, the temporary itself is destroyed, deleting the make_unique-created int.

The net result is delete p plus a useless new/delete cycle.

(It would be a declaration had it used parentheses rather than braces: std::unique_ptr<int> (p) = std::make_unique<int>(1); is exactly equivalent to std::unique_ptr<int> p = std::make_unique<int>(1);.)