Why is this an illegal constant expression?

Question

I am trying to preserve a variable so I can see its value while debugging optimized code. Why is the following an illegal constant expression?

   void foo(uint_32 x)
   {
       static uint_32 y = x;
       ...
   }

Answer 1

For your purpose you probably want this:

void foo(uint_32 x)
{
    static uint_32 y;
    y = x;
    ...
}

What you tried to do is an initialisation. What is done above is an assignment.

---

Maybe for your purpose this would be even more interesting:

static uint_32 y;
void foo(uint_32 x)
{
    y = x;
    ...
}

Now the variable y can easily be accessed by the debugger once the foo function is finished.