Why doesn't the standard make the '+=', '-=', etc. be defined only as
non-static member functions similar to the way that an 'operator=,
operator[], operator(), and operator->' must be non-static member
function, so that it will disallow 5 += obj;

In C++ Primer(4th Edition) by Stanley Lippman, in page 162, the
following is mentioned:

The general syntactic form of a compound assignment operator is

a op= b

where op= may be one of the following ten operators:
+= -= *= /= %= // arithmetic operators
<<= >>= &= ^= != // bitwise operators


In page 512 in this book, in the chapter 14 on Overloaded Operations
and Conversions, the following is mentioned:

Like assignment, the compound-assignment operators ordinarily ought to
be members of the class. Unlike assignment, they are not required to
be so and the compiler will not complain if a non-member compound-
assignment operator is defined.

I tried the following sample program:

#include <iostream>
#include <cstdlib>

using namespace std;

class Test
{
public:
Test(int arg = 10);
int val;
};

Test::Test(int arg) : val(arg)
{
cout << "one arg ctor called" << endl;
}

Test operator+=(Test lhs, Test rhs)
{
cout << "from operator+=" << endl;
return Test(lhs.val + rhs.val);
}

int main()
{
Test obj;

5 += obj;

return EXIT_SUCCESS;
}

This program compiles fine with both g++ and VC++ 2005 Express Edition
and both produced the following output:

one arg ctor called
one arg ctor called
from operator+=
one arg ctor called

My question:
Why doesn't the standard make the '+=', '-=', etc. be defined only as
non-static member functions similar to the way that an 'operator=,
operator[], operator(), and operator->' must be non-static member
function, so that it will disallow 5 += obj;

Is there any advantage for the compound-assignment operators(as per
the definition given in the beginning above) be allowed as non-member
functions ?