STACK FRAMES ANDFUNCTIONS

Game102 – Game Development

STACK OPERATIONSPUSH – Add an item to TOP

POP – Remove item from TOP

Consider a stack of books

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.

SAMPLE FUNCTIONint big ( int left, int right){ int largest;

if (left > right) largest = left; else largest = right; return largest;}

Stack Frame for“big”

big

• int left• int right

• int largest

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;}

Questions

1. Does main( ) create a stack frame when called?

2. What is on the stack frame for main( )?

3. How is the stack organized with multiple stack frames

Stack Organization

Another Functionint factorial( int n ){ int retval;

if ( n <= 1) retval = 1; else retval = n * factorial( n - 1 ); return ( retval );}

Factorial Flowchart

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”

Main Program

int main(){ cout << factorial(6) << endl; return 0;}

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

main( )

Fibonacci SeriesThe Fibonacci Sequence is the series of numbers:0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ...

We can solve this using this formulaN = 0… result is 0N = 1… result is 1N = n… result is f(n-1) + f(n-2)

Fibonacci Functionint 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;}

The main( ) Functionint main(){ cout << fib(4) << endl; return 0;}

The Stack (Maximum)fib(0)fib(1)fib(2)fib(1)fib(0)fib(1)fib(2)fib(3)fib(4)

main( )

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

An Iterative Solutionint 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;}

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

follow

Disadvantages of Iteration More difficult to write code Longer code

Passing by Valueint big ( int left, int right){ int largest;

if (left > right) largest = left; else largest = right; return largest;}

Passing by ValueA 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

Passing by ADDRESSvoid 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;}

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 integerif you set this pointer to an address, it can modify values AT that address

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 frameeither GLOBAL VARIABLES or variables on OTHER stack frames

Passing by REFERENCEvoid 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;}

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”