override #define value

No, you cannot override a #define constant with variable at run-time. The reason is simple: #define statements are processed by the compiler -- at compile-time, i.e. long before your program runs. And the compiler inserts those values as constants in your program code.

But you can do the following:

#define LIM 5

void MyFunction ()
{
    int x;

    if (!scanf ("%d", &x) != 1)
        x = LIM;
    ...    
}

As the other solutions have indicated, you can't change a value at runtime, but you can change the compile time definition of a macro during the compile process. For example:

void foo()
{
#define LIM 5
    int zork[LIM]; // defined as size 5

#undef LIM
#define LIM 42
    int blork[LIM];  // defined as size 42

#undef LIM
#define LIM 123

    // At this point, LIM is 123, but zork is still size 5 and blork is still
    // size 42. This does not change during the compile process or during runtime.
#undef LIM
    // At this point, LIM is undefined.
}

The answers given to your question and this link show why it is better to use a static const than a #define in this case.

http://stackoverflow.com/questions/1637332/static-const-vs-define[^]

However #define can be used to create a macro and this is quite legal though I would not recommend it.

#define LIM (x)

 int x = 5;

 cout << LIM << endl;

 x = 6;

  cout << LIM << endl;