Call of overloaded function is ambiguous

replace p.setval(0); with the following.

const unsigned int param = 0;
p.setval(param);

That way it knows for sure which type the constant 0 is.

The solution is very simple if we consider the type of the constant value, which should be "unsigned int" instead of "int".

Instead of:

setval(0)

Use:

setval(0u)

The suffix "u" tell the compiler this is a unsigned integer. Then, no conversion would be needed, and the call will be unambiguous.

The literal 0 has two meanings in C++.
On the one hand, it is an integer with the value 0.
On the other hand, it is a null-pointer constant.

As your setval function can accept either an int or a char*, the compiler can not decide which overload you meant.

The easiest solution is to just cast the 0 to the right type.
Another option is to ensure the int overload is preferred, for example by making the other one a template:

class huge
{
 private:
  unsigned char data[BYTES];
 public:
  void setval(unsigned int);
  template <class T> void setval(const T *); // not implemented
  template <> void setval(const char*);
};