functions-recall 1. 2 parameter passing & return void main() { int x=10, y=5; printf (“m1: x =...

$: Functions-Recall 1. 2 Parameter passing & return void main() { int x=10, y=5; printf (“M1: x = %d, y = %d\n”, x, y); interchange (x, y); printf (“M2:$
Functions-Recall

1

2

Parameter passing & returnvoid main()

{

int x=10, y=5;

printf (“M1: x = %d, y = %d\n”, x, y);

interchange (x, y);

printf (“M2: x = %d, y = %d\n”, x, y);

}

void interchange (int x, int y)

{ int temp;

printf (“F1: x = %d, y = %d\n”, x, y);

temp= x; x = y; y = temp;

printf (“F2: x = %d, y = %d\n”, x, y);

}

Output

3

Parameter passing & returnvoid main()

{

int x=10, y=5;

printf (“M1: x = %d, y = %d\n”, x, y);

interchange (x, y);

printf (“M2: x = %d, y = %d\n”, x, y);

}


{ int temp;

printf (“F1: x = %d, y = %d\n”, x, y);

temp= x; x = y; y = temp;

printf (“F2: x = %d, y = %d\n”, x, y);

}

M1: x = 10, y = 5

F1: x = 10, y = 5

F2: x = 5, y = 10

M2: x = 10, y = 5

Output

4

Parameter passing & returnint x=11,y=6;

void interchange (int a, int b);

int main()

{

{ int x=6,y=11;

interchange (x,y);

}

printf("x=%d y=%d",x,y);

}


{ int temp;

temp=x;

x=y;

y=temp;

}

Homework

5

How are function calls implemented? The following applies in general, the implementation

details of function call The system maintains a stack in memory

Stack is a last-in first-out structure Two operations on stack, push and pop

Whenever there is a function call, the activation record gets pushed into the stack

Activation record consists of the return address in the calling program, the return value from the function, and the local variables inside the function

6

Pop activation record, whenever the function returns

Activation record looks like:

Return Addr

Return Value

Local Variables

7

void main(){ …….. x = gcd (a, b); ……..}

int gcd (int x, int y){ …….. …….. return (result);}

Return Addr

Return Value

Local Variables

Before call After call After return

ST

AC

K

Activation record

8

void main(){ …….. x = ncr (a, b); ……..}

int ncr (int n, int r){ x=fact(n); y=fact(r); z=fact(n-r); p=x/(y*z) return (p);}

LV1, RV1, RA1

Before call Call fact ncr returns

int fact (int n){ ……… return (result);}

3 times

LV1, RV1, RA1

fact returns

LV1, RV1, RA1

LV2, RV2, RA2

Call ncr

3 times

Push activation record

9

Return Addr

Return Value

a, b (Local x Variables)

main


10

Return Addr

Return Value


Return Addr

Return Value

n, r, p (Local Variables)

main

nCr

Parameter passing

11

void main(){ …….. x = ncr (a, b); ……..}


LV1, RV1, RA1



3 times

LV1, RV1, RA1

fact returns

LV1, RV1, RA1

LV2, RV2, RA2

Call ncr

3 times


12

Return Addr

Return Value


Return Addr

Return Value

n, r (Local Variables)

Return Addr

Return Value

n, result (Local Variables)

result

x

main

nCr

Parameter passing

Return

fact

Pop activation record

13

Return Addr

Return Value

Local Variables

Return Addr

Return Value

Local Variables

main

nCr

14

void main(){ …….. x = ncr (a, b); ……..}


LV1, RV1, RA1



3 times

LV1, RV1, RA1

fact returns

LV1, RV1, RA1

LV2, RV2, RA2

Call ncr

3 times


15

Return Addr

Return Value

Local Variables

Return Addr

Return Value

Local Variables

Return Addr

Return Value

Local Variables

result

y

main

fact

nCr


16

Return Addr

Return Value

Local xVariables

Return Addr

Return Value p

Local Variables

main

Return

nCr


17

Return Addr

Return Value


main

18

Recursion

19

Recursion

A process by which a function calls itself repeatedlyEither directly.

X calls XOr cyclically in a chain.

X calls Y, and Y calls X

Used for repetitive computations in which each action is stated in terms of a previous result fact(n) = n * fact (n-1)

20

Contd.

For a problem to be written in recursive form, two conditions are to be satisfied: It should be possible to express the problem

in recursive form Solution of the problem in terms of solution of the

same problem on smaller sized dataThe problem statement must include a

stopping condition

fact(n) = 1, if n = 0 = n * fact(n-1), if n > 0

Stopping condition

Recursive definition

21

Examples:

Factorial:fact(0) = 1fact(n) = n * fact(n-1), if n > 0

Fibonacci series (1,1,2,3,5,8,13,21,….)fib (0) = 1fib (1) = 1fib (n) = fib (n-1) + fib (n-2), if n > 1

22

Factorial

long int fact (int n)

{

if (n == 1)

return (1);

else

return (n * fact(n-1));

}

23

Factorial Execution

long int fact (int n){ if (n = = 1) return (1); else return (n * fact(n-1));}

24

Factorial Execution fact(4)


25


if (4 = = 1) return (1);else return (4 * fact(3));


26





27






28

Factorial Execution

if (1 = = 1) return (1);

fact(4)





29

Factorial Execution

if (1 = = 1) return (1);

fact(4)



if (2 = = 1) return (1);else return (2 * fact(1)); 1


30

Factorial Execution

if (1 = = 1) return (1);

fact(4)




2


31

Factorial Execution

if (1 = = 1) return (1);

fact(4)




2


32

Factorial Execution

if (1 = = 1) return (1);

fact(4)




2

6


33

Factorial Execution

if (1 = = 1) return (1);

fact(4)




2

6

24


34

What happens for recursive calls? What we have seen ….

Activation record gets pushed into the stack when a function call is made

Activation record is popped off the stack when the function returns

In recursion, a function calls itselfSeveral function calls going on, with none of the

function calls returning back Activation records are pushed onto the stack continuously Large stack space required

35

Activation records keep popping off, when the termination condition of recursion is reached

We shall illustrate the process by an example of computing factorialActivation record looks like:

Return Addr

Return Value

Local Variables

36

Example:: main() calls fact(3)

int fact (n)

int n;

{

if (n = = 0)

return (1);

else

return (n * fact(n-1));

}

void main()

{

int n;

n = 3;

printf (“%d \n”, fact(n) );

}

37

TRACE OF THE STACK DURING EXECUTION

fact returns to main

RA .. main

-

n = 3

RA .. main

-

n = 3

RA .. fact

-

n = 2

RA .. main

-

n = 3

RA .. fact

-

n = 2

RA .. fact

-

n = 1

RA .. main

-

n = 3

RA .. fact

-

n = 2

RA .. fact

-

n = 1

RA .. fact

1

n = 0

RA .. main

-

n = 3

RA .. fact

-

n = 2

RA .. fact

1*1 = 1

n = 1

RA .. main

-

n = 3

RA .. fact

2*1 = 2

n = 2

RA .. main

3*2 = 6

n = 3

main calls fact

38

Look at the variable addresses (a slightly different program) !

void main()

{

int x,y;

scanf("%d",&x);

y = fact(x);

printf ("M: x= %d, y = %d\n", x,y);

}

int fact(int data)

{ int val = 1;

printf("F: data = %d, &data = %u \n

&val = %u\n", data, &data, &val);

if (data>1) val = data*fact(data-1);

return val;

}

4

F: data = 4, &data = 3221224528

&val = 3221224516

F: data = 3, &data = 3221224480

&val = 3221224468

F: data = 2, &data = 3221224432

&val = 3221224420

F: data = 1, &data = 3221224384

&val = 3221224372

M: x= 4, y = 24

Output

39

Fibonacci recurrence:

fib(n) = 1 if n = 0 or 1;

= fib(n – 2) + fib(n – 1)

otherwise;

int fib (int n) { if (n == 0 or n == 1)

return 1; [BASE] return fib(n-2) + fib(n-1) ; [Recursive] }

Fibonacci Numbers

40

fib (5)

fib (3) fib (4)

fib (1)

fib (2)fib (1) fib (2)

fib (0)

fib (3)

fib (1)

fib (1) fib (2)

fib (0)

fib (0) fib (1)


fib(n) = 1 if n = 0 or 1;

= fib(n – 2) + fib(n – 1)

otherwise;

int fib (int n) { if (n == 0 || n == 1)

return 1; return fib(n-2) + fib(n-1) ;}

41

fib (5)

fib (3) fib (4)

fib (1)


fib (0)

fib (3)

fib (1)

fib (1) fib (2)

fib (0)

fib (0) fib (1)

1 1 1 1 1

1

11


fib(n) = 1 if n = 0 or 1;

= fib(n – 2) + fib(n – 1)

otherwise;

int fib (int n) { if (n == 0 || n == 1)


fib.c

42

fib (5)

fib (3) fib (4)

fib (1)


fib (0)

fib (3)

fib (1)

fib (1) fib (2)

fib (0)

fib (0) fib (1)

1 1 1 1 1

1

11

1 2 2

211111

3

3

5

8

1 1


fib(n) = 1 if n = 0 or 1;

= fib(n – 2) + fib(n – 1)

otherwise;

int fib (int n) { if (n==0 || n==1)


43

Mergesort

44

Basic Idea Divide the array into two halves Sort the two sub-arrays Merge the two sorted sub-arrays into a single

sorted array Step 2 (sorting the sub-arrays) is done

recursively (divide in two, sort, merge) until the array has a single element (base condition of recursion)

45

Merging Two Sorted Arrays

3 4 7 8 9

2 5 7

2

3 4 7 8 9

2 5 7

2 3

3 4 7 8 9

2 5 7

2 3 4

3 4 7 8 9

2 5 7

Problem: Two sorted arrays A and B are given. We are required to

produce a final sorted array C which contains all elements of A and B.

46

2 3 4 5

3 4 7 8 9

2 5 7

2 3 4 5 7

3 4 7 8 9

2 5 7

2 3 4 5 7 7

3 4 7 8 9

2 5 7

2 3 4 5 7 7 8

3 4 7 8 9

2 5 7

47

Merge Code

2 3 4 5 7 7 8 9

3 4 7 8 9

2 5 7

void

merge (int A[], int B[], int C[], int m,int n)

{

int i=0,j=0,k=0;

while (i<m && j<n)

{

if (A[i] < B[j]) C[k++] = A[i++];

else C[k++] = B[j++];

}

while (i<m) C[k++] = A[i++];

while (j<n) C[k++] = B[j++];

}

48

Merge Sort: Sorting an array recursively

void mergesort (int A[], int n)

{

int i, j, B[max];

if (n <= 1) return;

i = n/2;

mergesort(A, i);

mergesort(A+i, n-i);

merge(A, A+i, B, i, n-i);

for (j=0; j<n; j++) A[j] = B[j];

free(B);

}

functions-recall 1. 2 parameter passing & return void main() { int x=10, y=5; printf (“m1: x =...

Documents

y interchange x

int gcd int x

y temp

returnint x

factn y

void interchange int

int ncr int n

y printf m2