Question about ssa-form in gcc

[Eugen Wagner]

Hi,
noticed, that after converting into ssa-form, there are still
variables, which are not wrapped into SSA_NAME.

Assigment to scalar variable:
int a;
a  = 100;
The variable a is not wrapped into SSA_NAME a_1.

b = foo();
is converted into
 b.0_3 = foo ();
 # .MEM_25 = VDEF <.MEM_24>
 b = b.0_3;

Question:
What is the purpose of this behavour? In classical ssa-representation
all variables should be numbered.
I guess it has somethink to do with memory-ssa and alias analysis.
Could anybody confirm or negate?

A simple program and the cut-off of the  dump after putting into ssa follows.

int foo(){
  return 100;
}
int main(int argc, char **args){
  int a,b,d, *pa;
  a=10;
  b=20;
  if(argc>3){
    b=foo();
    pa=&a;
    b=*pa + 3;
  }
  else{
    a+=1;
    pa=&b;
  }
  *pa += 100;
  b = a + b;
  a++;
  printf("*pa=%d\n",*pa+100);
  return 0;

}


;; Function foo (foo)

foo ()
{
  int D.2572;

<bb 2>:
  D.2572_1 = 100;
  return D.2572_1;

}

main (int argc, char * * args)
{
...
<bb 2>:
  # .MEM_23 = VDEF <.MEM_22(D)>
  a = 10;
  # .MEM_24 = VDEF <.MEM_23>
  b = 20;
  if (argc_2(D) > 3)
    goto <bb 3>;
  else
    goto <bb 4>;

<bb 3>:
  b.0_3 = foo ();
  # .MEM_25 = VDEF <.MEM_24>
  b = b.0_3;
  pa_4 = &a;
  # VUSE <.MEM_25>
  D.2561_5 = *pa_4;
  b.1_6 = D.2561_5 + 3;
  # .MEM_26 = VDEF <.MEM_25>
  b = b.1_6;
  goto <bb 5>;
...
}