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

Question

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; 

Answer 1

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.