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Problems Problem 1 In this problem, we will use and create function that utilize function pointers. The file ’call back.c’ contains an array of records consisting of a fictitious class of celebrities. Each record consists of the firstname, lastname and age of the student. Write code to do the following: • Sort the records based on first name. To achieve this, you will be using the qs o r t () function provided by the standard library: void qsort(void∗ arr,int num,int size,int (∗fp)(void∗ pa,void∗pb)). The function takes a pointer to the start of the array ’arr’, the number of elements ’num’ and size of each element. In addition it takes a function pointer ’fp’ that takes two arguments. The function fp is used to compare two elements within the array. Similar to strcmp(), it is required to return a negative quantity,zero or a positive quantity dependeing on whether element pointed to by ’pa’ is ”less” than, equal to or ”greater” the element pointed to by ’pb’. You are required to write the appropriate callback function.

• Now sort the records based on last name. Write the appropriate callback function.

• The function void apply (...) iterates through the elements of the array calling a function for each element of the array. Write a function i s o l d e r ( ) that prints the record if the age of the student is greater 20 and does nothing otherwise.

Problem 2 A useful data structure for doing lookups is a hash table. In this problem, you will be imple menting a hash table with chaining to store the frequency of words in a file. The hash table is implemented as an array of linked lists. The hash function specifies the index of the linked list to follow for a given word. The word can be found by following this linked list. Optionally, the word can be appended to this linked list. You will require the code file ’hash.c’ and data file ’book.txt’ for this problem. You are required to do the following • The function lookup() returns a pointer to the record having the required string. If not found it returns NULL or optionally creates a new record at the correct location. Please complete the rest of the code.

• Complete the function cl e ar ta bl e ( ) to reclaim memory. Make sure each call to malloc() is matched with a f r e e ( ) programminghomeworkhelp.com

Solutions Answer: Here’s one possible implementation: # in c lu d e < s t d i o . h> # in c l u d e < s t r i n g . h> # i n c lu d e < s t d l i b . h> # d e fi n e MAX STUDENTS 10 s t r u c t student { ch ar fname [ 1 0 0 ] ; char lname [ 1 0 0 ] ; i nt year ; i nt age ; } ; s t r u c t st udent c l a s s [ ] = { " Sean " , " Penn" , 2 , 2 1 , " Sean " , " Conner y " , 4 , 2 5 , " Angel i na " , " J ol i e " , 3 , 2 2 , " Mer yl " , " St r eep " , 4 , 2 9 , " Robi n " , " Wi l l i ams " , 3 , 3 2 , " Bi l l " , " Gat es " , 3 , 1 7 , " J odi e " , " Fos t er " , 4 , 2 5 , " J ohn " , " Tr avol t a " , 1 , 1 7 , " I s aac " , " Newt on " , 2 , 1 9 , " Sar ah " , " Pal i n " , 2 , 19 } ; /∗ @ funct i on co m p ar e f i r st n am e @desc compares f i r s t name o f two r e c o r d s . ∗/ i nt co mpare f irst name ( const void ∗ a , const void ∗ b) { s t r u c t student ∗ s a = ( s t r u c t student ∗) a ;

struct

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} /∗ @ funct ion compare lname name @desc compares l a s t name o f two r e c o r d s . ∗/ i n t compar e last name ( c on st v oid ∗ a , c on st v oid ∗ b) { s t r u c t student ∗ s a= ( s t r u c t student ∗) a ; s t r u c t student ∗ s b=( s t r u c t student ∗) b ; ret ur n strcmp ( sa−>lname , sb−>lname ) ; } /∗ ! @ funct ion apply @desc a p p l i e s ∗/ void apply ( s t r u c t student ∗ sarr , i nt nrec , void ( ∗ fp ) ( void ∗ prec , void∗ arg ) , void ∗ arg ) { i nt i =0; for ( i = 0; i <nr ec ; i + + ) { }

}

/ ∗ c a l l b a c k ∗/

fp (& s a r r [ i ] , ar g ) ;

/∗ @ funct ion p r i n t r e c @desc p r i n t s st udent r e co r d ∗/ v oid p r i n t r e c ( v oid ∗ pr ec , v oid ∗ ar g ) {

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}

s t r u c t student ∗ pstud=( s t r u c t student ∗) prec ; p r i n t f ( "%-20 s %-20 s %2 d %2 d\n" , pstud−>fname , pstud−>lname , pstud−>ye ar , pstud−>age ) ;

/∗ @ funct ion i s o l d e r @desc p r i n t s st udent r e co r d ∗/ v oid i s o l d e r ( v oid ∗ prec , v oid ∗ ar g ) { i n t ∗ age= ( i n t ∗) ar g ; s t r u c t student ∗ pstud=( s t r u c t student ∗) prec ; i f ( pstud−>age < ∗ age ) r et u r n ; / ∗ do not hi n ∗/ else p r i n t f ( "%-20 s %-20 s %2 d %2 d\n" , pstud−>fname , pstud−>lname , pstud−>ye ar , pstud−>age ) ; } i n t main ( ) { i nt n st u de nt s = s i ze o f ( c l a s s ) / s i ze o f ( s t r u c t student ) ; i nt age ; put s ( " Raw r ec or ds : " ) ; puts ( " - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - " ) ; apply ( c l a s s , nst udent s , p r i n t r e c ,NULL ) ; / ∗ s o r t based on f i r s t name∗/ put s ( " Sor t ed by f i r s t name: " ) ; puts ( " - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - " ) ;

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q s o r t ( c l a s s , nstudents , s i z e o f ( apply ( c l a s s , ns tudents , p r i n t r e first name); / ∗ s o r t based on l a s t name ∗/ puts ( " -------------------------------" ) ; q s o r t ( c l a s s , nstudents , s i z e o f ( r i n t r e c ,NULL) ;

struct s tu den t ) , c o m p a r e c ,NULL) ; puts ( " Sorted by last name:" ) ; struct s tu den t ) , compar e apply ( c l a s s , ns tudents , p

last name ) ; / ∗ p r i n t p e o p l e o l d e r than 20 ∗/ puts ( " People older than 20: " ) ; puts ( " -------------------------------" ) ; age =20; apply ( c l a s s , ns tudents , i s o l d e r ,& age ) ; return 0 ; } Answer: one possible implementation is shown below: # i n c lu d e < s t d i o . h> # i n c lu d e < s t d l i b . h> # in c lu d e < s t r i n g . h> # d e fi n e MAX BUCKETS 1000 # d e fi n e MULTIPLIER 31 # d e fi n e MAX LEN 100 s t r u c t wordrec { char∗ word ; unsigned long count ; s t r u c t wordr ec ∗ nex t ; } ; programminghomeworkhelp.com

s t r u c t wordr ec ∗ w al l o c ( c on st char ∗ s t r ) { s t r u c t wordrec ∗ p= ( s t r u c t wordrec ∗) malloc ( s i ze o f ( s t r u c t wordrec ) ) ; i f ( p!= NULL ) { p−>co unt = 0; p−>word= s t r dup ( s t r ) ; / ∗ c r e a t e s a d u p l i c a t e ∗/ p−>n ext=NULL; } retu r n p ; } / ∗ hash buck et ∗/ s t r u c t wordr ec ∗ t a b l e [MAX LEN ] ; /∗ @ funct ion @desc h a s h s t r i n g pr oduces hash code f o r a s t r i n g ll

m u l t i p l i e r s 31 , 35 have been found t o work w e

∗/ un signed lon g h a s h s t r i n g ( c on st ch ar ∗ s t r ) { u n sign ed lon g hash = 0; w h ile ( ∗ s t r ) { hash= hash ∗MUL T IPL IE R+ ∗ s t r ; s t r + + ; } r et u r n hash%MAX BUCK ET S ; } /∗ @ funct ion look up @desc r e t u r n s a p o i n t e r t o t he word or c r e a t e s i t i f re q u i re d ∗/ s t r u c t wordr ec ∗

look up ( c on st char ∗ s t r , i n t c r e a t e )

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s t r u c t wordr ec ∗ c u r r=NULL ; for ( cur r= wp ; cur r != NULL ; cu r r= cur r −>n ex t ) i f ( strcmp ( cur r −>word , s t r )= = 0) / ∗ found ∗/ { ret ur n curr ; } / ∗ not found ∗/ i f ( c r e a te ) { c u r r = ( s t r u c t wor drec ∗) m al l oc ( s i z e o f ( s t r u c t curr wor drec ) ) ; −>word= s t r dup ( s t r ) ; cur r −>co unt = 0; / ∗add t o f r o n t ∗/ cur r −>n ex t= t a b l e [ hash ] ; t a b l e [ hash ]= cur r ; } ret ur n curr ; } /∗ @ funct i on c l e a r t a b l e ( ) @desc r e c l a i m s memory ∗/ v oid c l e a r t a b l e ( ) { s t r u c t wordrec ∗ wp=NUL L , ∗ p=NULL ; i nt i =0; for ( i = 0; i <MAX BUCK ETS ; i + + ) { wp= t a b l e [ i ] ; w h ile ( wp) { p= wp ; wp= wp−>n ex t ; f r e e ( p−>word ) ; f r e e ( p ) ; programminghomeworkhelp.com

{

un sign ed lon g hash= h a s h s t r i n g ( s t r ) ; s t r u c t wordrec ∗ wp= t a b l e [ hash ] ; }}}

i n t main ( i n t ar gc , ch ar ∗ argv [ ] ) { FILE∗ fp=fopen ( " book. t x t " , " r " ) ; ch ar word [ 1 0 2 4 ] ; / ∗ bi g enough ∗/ s t r u c t wor drec ∗ wp=NULL ; i nt i =0; memset ( t abl e , 0 , s i z e o f ( t a b l e ) ) ; / ∗ r ead from i nput ∗/ w h ile ( 1 ) { i f ( f s c a n f ( fp , " % s " , word ) ! = 1 ) br eak ; doesn ’ t e x i s t ∗/ wp= lo ok up ( word , 1 ) ; / ∗ c r e a t e i f wp−>co u nt + + ; } f c l o s e ( fp ) ; /∗ p r i n t a l l words have fr equency >100 ∗/ for ( i = 0; i <MAX BUCK ETS ; i + + ) { for ( wp= t a b l e [ i ] ; wp!= NULL ; wp= wp−>n ex t ) {

i f ( wp−>count >1000) { p r i n t f ( " % s - - >% l d \ n " ,wp−>word , wp−>co unt ) ; }}

cleartable (); r et u r n 0 ; programminghomeworkhelp.com