Why is the destructor of a function parameter not called at the end of the function, but at the end of the full expression making the function call?

Consider the following example:

#include <iostream>

struct A
{
    int n = 0;
    A() { std::cout << "A()" << std::endl; }
    A(const A&) { std::cout << "A(const A&)" << std::endl; }
    ~A() { std::cout << "~A()" << std::endl; }
};

int f(A a) { std::cout << "f()" << std::endl; return 1; }

int main()
{
    A a;
    std::cout << f(a) << " was returned from f(A)" << std::endl;
    std::cout << "leaving main..." << std::endl;
}

I expect the output of this program to be:

A()
A(const A&)
f()
~A()
1 was returned from f(A)
leaving main...
~A()

because when the function returns, its parameters should have already been destroyed. For example, if we have the following function:

A* g(A a) { std::cout << "g()" << std::endl; return &a; }

Dereferencing the returned pointer is undefined behavior:

std::cout << g(a)->n << std::endl; // undefined behavior

But the output is:

A()
A(const A&)
f()
1 was returned from f(A)
~A()
leaving main...
~A()

I'm looking for an explanation of this behavior.