Can formal parameters inside the function not be dropped even if the function returns until the caller statement ends?

I have tried some compilers and some C++ standard versions. This code may have a deadlock (the comments point out that this is UB) and can be reproduced at least under gcc8 and C++11. I use "Compiler Explorer" to check out the assembly of this code. And I find that the argument q of function foo1 is not dropped until the caller statement in main ends.

https://godbolt.org/z/vaes53z44

#include <iostream>
#include <memory>
#include <mutex>

std::mutex mtx;

int foo1(std::unique_lock<std::mutex> q) {
    std::cout << "foo1" << std::endl;
    return 0;
}

void foo2(int _v) {
    std::unique_lock<std::mutex> q(mtx);
    std::cout << "foo2" << std::endl;
}

int main() {
    foo2(foo1(std::unique_lock<std::mutex>(mtx)));
    return 0;
}

Yes, parameters may live until the end of the enclosing full-expression. In this case, the full-expression is foo2(foo1(std::unique_lock<std::mutex>(mtx))).

[expr.call]/6, emphasis mine:

It is implementation-defined whether a parameter is destroyed when the function in which it is defined exits or at the end of the enclosing full-expression; parameters are always destroyed in the reverse order of their construction.