What's the purpose of using a union with only one member?

When I was reading seastar source code, I noticed that there is a union structure called tx_side which has only one member. Is this some hack to deal with a certain problem?

FYI, I paste the tx_side structure below:

union tx_side {
    tx_side() {}
    ~tx_side() {}
    void init() { new (&a) aa; }
    struct aa {
        std::deque<work_item*> pending_fifo;
    } a;
} _tx;

Because tx_side is a union, tx_side() doesn't automatically initialize/construct a, and ~tx_side() doesn't automatically destruct it. This allows a fine-grained control over the lifetime of a and pending_fifo, via placement-new and manual destructor calls (a poor man's std::optional).

Here's an example:

#include <iostream>

struct A
{
    A() {std::cout << "A()\n";}
    ~A() {std::cout << "~A()\n";}
};

union B
{
    A a;
    B() {}
    ~B() {}
};

int main()
{
    B b;
}

Here, B b; prints nothing, because a is not constructed nor destructed.

If B was a struct, B() would call A(), and ~B() would call ~A(), and you wouldn't be able to prevent that.

In simple words, unless explicitly assigned/initialized a value the single member union does not initialize the allocated memory. This functionality can be achieved with std:: optional in c++17.