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.

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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.

Answer 2

"Why is the following an illegal constant expression?"

Because static variables have to be initialized with a value known at compile-time, while x is only determined at run-time.

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Note that this use of static is meant for to keeping the variable with its stored value between different calls to foo() alive (existing in memory) - Means the object won´t get destroyed/ deallocated after one single execution of the function, as it is the case with function-local variables of the storage class automatic.

It wouldn´t make sense to create and initialize a static variable at each function call new.