Appendix
+ Solved Question Papers
C Programming and Data Structures May/June 2007 SET- 1 1.
(a) Write a program to determine and print the sum of the following harmonic series for a given value of n:1+1/2+1/3+1/4+1/5 +1/n. #include<stdio.h> #include<conio.h> main() { int i,n; float sum=0; clrscr(); printf( enter the value of n ); scanf( %d ,&n); for(i=1;i<=n;i++) sum+=1/(float)i; printf( the value of the given series is %f\n ,sum); getch(); }
(b) What are the logical operators used in C ? Illustrate with examples. Logical operators used in C language are && logical AND || logical OR ! logical NOT
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C.2 Solved Question Papers Truth table for logical operators: A T T F
F
B T
A&&B T
A||B T
!A F
F
F
T
F
T F
F F
T F
T T
Logical AND is true only when all the inputs are true and logical OR is true if any one of the inputs is true. Logical NOT is a unary operator, it is true when input is false and viceversa. Sample program for logical operators:
2.
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#include<stdio.h> main() { int a,b,c,nc=0,nw=0; char ch; FILE *fptr; printf( enter the values of a,b,c\n ); scanf( %d%d%d ,&a,&b,&c); if((a>b)&&(b>c)) /* logical AND */ printf( a is the largest value ); fptr=fopen( input.c , r ); while(!foef(fptr)) /* logical NOT */ { ch=getc(fptr); if(ch== ||ch== \t ) /* logical OR */ nw++; else nc++; } fclose(fptr); printf( number of words =%d\n ,nw); printf( number of characters =%d\n ,nc); } getch(); }
(a) What is a preprocessor directive? The preprocessor is a program that processes the source program before it is passed on to the compiler. The program typed in the editor is the source to the preprocessor. The preprocessor directives are generally initialized at the beginning of the program. It begins with a symbol #(hash). Example : #define pi 3.14
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(b) Distinguish between function and preprocessor directive. Function
Preprocessor Directive
Function is a block of statements enclosed within braces.
Preprocessor directives are mostly single line statements usually used to define symbolic constants. Compiler replaces the code when it is encountered. Preprocessor can neither accept nor return values. Example: #define pi 3.14
Compiler moves to the function where it is located when a function is invoked. Functions can accept and return values from and to the calling function. Example: sum(int a, int b) { return a+b;}
(c) What is the significance of conditional compilation? Conditional compilation allows the user to compile the portions of the program based on certain conditions. The directives used for conditional compilation are #if, #else, #endif, etc. The syntax for #if directive is #ifdef <identifier> { statement1; statement2; . . } #else { statement x; statement y; . . } #endif
( d) How is the unedifining of a pre-defined macro done? A preprocessor directive that is defined using #define can be undefined using #undef directive. Syntax : #undef identifier It is useful when we do not want to allow the use of macros in any portion of the program. Example: #include<stdio.h> #define wait getch() main() { int a; clrscr(); #undef wait getch()
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C.4 Solved Question Papers
3.
for(a=1;a<=10;a++) { printf( %d\t ,a); wait;/* returns an error message since wait is undefined */ } }
(a) What is a pointer? How is a pointer initiated? Give an example. A pointer is a variable which stores the address of another variable. The syntax for declaring a pointer is data type * pointer_name; Example: int *ptr; char *cptr; float *p;
Once a pointer variable is declared, it can be assigned the address of a variable using assignment statement. Example: int a, *ptr; ptr=&a; & is called address of operator. ptr is assigned the address of a. The pointer and variable must be of the same data type. (b) State whether each of the following statements is true or false. Give reasons. (i) An integer can be added to a pointer. True. Example: ptr=ptr+1; here the pointer is incremented depending on its data type. If it is an integer then the scale factor is 2 bytes, character 1 byte, and so on. (ii) A pointer can never be subtracted from another pointer. False. Because C supports pointer arithmetic, by which a pointer can be subtracted or added to another pointer. (iii) When an array is passed as an argument to a function, a pointer is passed. True. Because the array name is itself a constant pointer to that array. (iv) Pointers cannot be used as formal parameters in headers to function definitions. False. C supports call by reference in which the formal parameters are pointers. (c) If m and n have been declared as integers and p1 and p2 as pointers to integers, then find out the errors, if any, in the following statements. (i) p1=&m; The given statement is correct. (ii) p2 = n; The given statement is an error as the compiler cannot convert int to int *. (iii) m=p2-p1; The given statement is correct as C supports pointer arithmetic. (iv) *p1=&n; The given statement is not correct as the compiler cannot convert int* to int.
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(a) What are bit fields? What are its advantages? What is its syntax? Bit fields are the special features provided in C to change the order of allocation of memory from bytes to bits. Using bit fields we can specify the exact number of bits required for the storage of values. The number of bits required for a variable is specified by a non-negative integer preceded by a colon. Advantages: · Changes order of allocation of memory storage. · Minimizes the memory wastage. · Memory can be used more effectively and efficiently. Syntax for using bit fields: unsigned variable : size; Example: unsigned pass:1; unsigned fail:0; (b) Write a C program to store the information of vehicles, use bit fields to store the status information. Assume the vehicle object consists of type, fuel and model member fields. Assume appropriate number of bits for each field. #include<stdio.h> #include<conio.h> #define petrol 1 #define diesel 2 #define two_wheeler 3 #define four_wheeler 4 #define old 5 #define new 6 main() { struct vehicle { unsigned type:3; unsigned fuel:2; unsigned model:3; }v; v.type=four_wheeler; v.fuel=diesel; v.model=new; printf( Type of vehicle :%d\n,v.type); printf( Fuel :%d\n ); printf( Model :%d\n ); getch(); } OUTPUT: Type of vehicle :4 Fuel :2 Model :6
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C.6 Solved Question Papers 5. Write a C program to read a text file and to count (a) number of characters (b) number of words (c) number of sentences and write in an output file #include<stdio.h> #include<conio.h> main () { FILE *fptr; int nl=0,nw=0,nc=0; char ch; clrscr(); fptr=fopen( input.c , r ); if (fptr!=NULL) { while (!foef(fptr)) { ch=getc (fptr); if (ch== ||ch== \t ) nw++; else if(ch== \n ) nl++; else nc++; } fclose (fptr); fptr=fopen ( output.c , w ); fprintf (fptr, number of lines =%d\n , nl); fprintf (fptr, number of words =%d\n , nw); fprintf (fptr, number of characters =%d\n , nc); } else printf( the input file does not exist\n ); getch(); }
6. Declare a circular queue of integers such that F points to the front and R points to the rear. Write functions (a) to insert an element into the queue (b) to delete an element from the queue #include<stdio.h> #define size 20 int cq[size],F=0,R=0,count=0; main() { int a,ch; clrscr();
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Solved Question Papers while(1) { printf( main menu\n ); printf( 1.insertion\t2.deletion\t3.display\t4.exit\n ); printf( enter your choice\n ); scanf( %d ,&ch); switch(ch) { case 1:
case 2: case 3: case 4: default: } getch();
printf( enter the value to be inserted\n ); scanf( %d ,&a); insert(a); break; delete(); break; display(); break; exit(0); printf( wrong choice\n ); }
/* insertion function */ insert(int x) { if(F==R&&count==size-1) { printf( circular queue is full\n ); return; } cq[R]=x; R=(R+1)%size; } /* deletion function */ delete() { if(F==R&&count==0) { printf( circular queue is empty\n ); return; } printf( the deleted element is %d\n ,cq[F]);
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C.8 Solved Question Papers F=(F+1)%size; } /* display function */ display() { int i; if(F==R&&count==0) { printf( circular queue is empty\n ); return; } for(i=F;i!=R;i=(i+1)%size) printf( %d ,cq[i]); }
7. Write a function in C to form a list containing the intersection of the elements of two lists. Intersection() { if(first1==NULL) { printf( list is empty. So no elements in common\n ); return; } else if(first2==NULL) { printf( list is empty. So no elements in common\n ); return; } else { ptr1=first1; ptr2=first2; while(ptr1->next!=NULL) { ptr2=first2; while(ptr2->next!=NULL) { if(ptr1->data==ptr2->data) { if(first3==NULL) { first3=(node *)malloc(sizeof(node)); first3->data=ptr1->data; first3->next=NULL; } else {
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Solved Question Papers ptr3=first3; while(ptr3->next!=NULL) ptr3=ptr3->next; fresh=(node *)malloc(sizeof(node));
fresh->data=ptr1->data; ptr3->next=fresh;
}/*
8.
fresh->next=NULL; }/* close of else }/* close of if */ }/* close of inner while loop */ }/* close of outer while loop */ close of function intersection */
*/
(a) Write a C program to sort given integers using partition exchange sort. Partition exchange sort is also called quick sort. /* main function */ #include<stdio.h> #include<conio.h> main() { int i,n,a[20]; clrscr(); printf( enter the size of array\n ); scanf( %d ,&n); printf( enter the elements into array\n ); for(i=0;i<n;i++) scanf( %d ,a+i); quicksort(a,0,n-1); prints( elements after sorting are\n ); for(i=0;i<n;i++) printf( %d\t ,*(a+i)); getch(); } /* QUICK SORT FUNCTION */ quicksort(int *k,int lb,in tub) { int pivot,i,j; if(lb<ub) { i=lb; j=ub; pivot=i; while(i<j) { while((*(k+i)<=*(k+pivot))&&(i<ub)) i++; while(*(k+j)>*(k+pivot)) j ;
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C.10 Solved Question Papers if(i<j) swap(k+i,k+j); } swap(k+j,k+pivot); quicksort(k,lb,j-1); quicksort(k,j+1,ub); } return; } /* Function for swapping swap(int *a, int *b) { int temp; temp=*a; *a=*b; *b=temp; }
*/
(b) Derive the time complexity of the following partition exchange sort. n
CA(n)=1/n å K = 1 {CA(k 1)+ CA(n k)}+(n+1) nCA(n)=n(n+1)+
n
å K = 1 {CA(k 1)+ CA(n k)}
=n(n+1)+{CA(0)+ CA(n 1)+ CA(1)+ CA(n 2) + + CA(n 1) + CA(0)} nCA(n)=n(n+1)+2{CA(0)+ CA(1)+ ..+ CA(n 1)} (1) replace n with (n 1) (n 1)CA(n 1)=n(n 1)+2{CA(0)+ CA(1)+ ..+ CA(n 2)} .(2) Subtract (2) from (1) nCA(n) (n 1)CA(n 1)=n2+n n2+n+2 CA(n 1) nCA(n)=2n+2 CA(n 1)+ (n 1)CA(n 1) nCA(n)=2n+ CA(n 1)[n 1+2] nCA(n)=2n+ CA(n 1)[n+1] divide on both sides by n(n+1) (CA(n)/(n+1))=(2/(n+1))+ (CA(n 1)/n) (CA(n)/(n+1))=(2/(n+1))+ 2/n+(CA(n 2)/(n 1)) =2/(n+1)+2/n+2/(n 1)+2/(n 2)+..+2/4+ CA(2)/3+2/3+ CA(1)/2 =2å
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n +1 K =3
(1/k)
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Solved Question Papers
=2ò
n +1
3
1/kdk n +1
=2[logk] 3
(CA(n)/(n+1))=2[log(n+1) log3] CA(n)=2(n+1)log(n+1) 2(n+1)log3 CA(n)=2nlog(n+1)+2 log(n+1) 2nlog3 2 log3 CA(n) »2nlog(n+1) CA(n)=O(nlogn)
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C.11