1. Stack Frames and Functions
Game102 – Game Development

2. Stack Operations
PUSH – Add an item to TOP POP – Remove item from TOP Consider a stack of books

3. Stack frame
Every function call will cause the creation of a stack frame
Stack frames contain LOCAL variables and ARGUMENTS to functions
Stack frames are CREATED when functions are called and DESTROYED when functions return.

4. Sample function int big ( int left, int right) { int largest;
if (left > right)
largest = left;
else
largest = right;
return largest; }

5. big int left int right int largest
Stack Frame for “big”
big
int left
int right
int largest

6. Main Program int main() { int val1; int val2; int val3;
cout << "Enter 3 values: " ;
cin >> val1 >> val2 >> val3;
cout << "Largest value is " << big(val1, big(val2,val3)) << endl;
return 0; }

7. Questions Does main( ) create a stack frame when called?
What is on the stack frame for main( )?
How is the stack organized with multiple stack frames

8. Stack Organization

9. Another Function int factorial( int n ) { int retval; if ( n <= 1)
else
retval = n * factorial( n - 1 );
return ( retval ); }

10. Factorial Flowchart

11. A Function the Calls Itself
Called “recursion”
Must have an exit path (typically a condition that causes it to return a value rather than call itself)
Is dangerous if not used properly
Works using the stack and “stack frames”

12. Main Program int main() { cout << factorial(6) << endl;
return 0; }

13. The Stack (Maximum)
factorial(1) factorial(2) factorial(3) factorial(4) factorial(5) factorial(6) main( )

14. Fibonacci Series
The Fibonacci Sequence is the series of numbers: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ... We can solve this using this formula N = 0… result is 0 N = 1… result is 1 N = n… result is f(n-1) + f(n-2)

15. Fibonacci Function
int fib( int n ) { int retval; if (n == 0) retval = 0; else if (n == 1) retval = 1; else retval = fib(n-1) + fib(n-2); return retval; }

16. The main( ) Function
int main() { cout << fib(4) << endl; return 0; }

17. fib(0) fib(1) fib(2) fib(3) fib(4) main( )
The Stack (Maximum)
fib(0) fib(1) fib(2) fib(3) fib(4) main( )

18. Why the strange display?
Every “n” value causes 2 RECURSIVE paths
N = 4 generates recursive N = 3 and N = 2
It is VERY bad when bigger numbers are used
BOTTOM LINE – be very careful if you used recursion – it is not always fast
Fast to write code – not always best

19. An Iterative Solution
int fib( int n ) { int retval; int i; int lastval = 0; int last2val = 1; int sum; if (n <= 0) retval = 0; else if (n == 1) retval = 1; else for (i = 1; i<=n ; i++) sum = lastval + last2val; last2val = lastval; lastval = sum; } retval = sum; return retval;

20. Advantages of Iteration
ONE stack frame call
VERY fast
Uses explicit loop so it should be easier to follow

21. Disadvantages of Iteration
More difficult to write code
Longer code

22. Passing by Value
int big ( int left, int right) { int largest; if (left > right) largest = left; else largest = right; return largest; }

23. Passing by Value
A COPY of the calling argument gets placed on the stack frame for the function call. If you call “big( )” with val1 and val2… big(val1,val2) a COPY of val1 and a COPY of val2 are placed into the stack frame of “big ( )” with “left” set to the val1 copy and “right” set to the val2 copy

24. Passing by ADDRESS
void swap ( int *pleft, int *pright) { int tmp; tmp = *pleft; *pleft = *pright; *pright = tmp; } int main() int val1 = 23; int val2 = 12; swap( &val1, &val2); cout << "val1 = " << val1 << " and val2 = " << val2 << endl; return 0;

25. Explanation
“&val1” means ADDRESS OF val1 this is the memory location of “val1”
“int *pleft” means the VALUE AT pleft is an int “pleft” is consider a POINTER to an integer if you set this pointer to an address, it can modify values AT that address

26. Explanation On a function call, a stack frame is created
When the function returns, the stack frame is destroyed
If you play with argument values and local variable in a function, they disappear after the function completes
If you want to save information for after the completion of the function, you must modify values OUTSIDE the stack frame either GLOBAL VARIABLES or variables on OTHER stack frames

27. Passing by REFERENCE
void swap ( int &left, int &right) { int tmp; tmp = left; left = right; right = tmp; } int main() int val1 = 23; int val2 = 12; swap( val1, val2); cout << "val1 = " << val1 << " and val2 = " << val2 << endl; return 0;

28. Explanation
Specifying “int &left” as an ARGUMENT makes “left” a reference to the variable that was passed (in this case “val1”)
Modifying “left” will modify “val1”