F.E. Sem. II. Structured Programming Approach

Transcription

1 F.E. Sem. II Structured Programming Approach Time : 3 Hrs.] Mumbai University Examination Paper Solution - May 13 [Marks : 80 Q.1(a) Explain the purpose of following standard library functions : [3] (i) Floor ( ) (ii) ceil ( ) (iii) sqrt ( ) (A) Rounding is done to the nearest integer. Actual rounding is done with respect to 0.5. C provides function for either rounding up or rounding down : (i) Floor ( ) is used for rounding down. [1 mark] (ii) ceil ( ) is used for rounding up. [1 mark] (iii) sqrt ( ) is used to find the square root of the given number. [1 mark] Q.1(b) Explain how problem is defined with a help of suitable example. [4] (A) The problem is defined with respect to inputs, outputs and the steps required for processing. The input along with its type is identified. The formula required for finding the answer is also identified with all its substeps. The type of the output to be produced is also finalized. [3 marks] e.g. To find the area of circle : [1 mark] Inputs : radius of a circle Type : real Output : area Type : real Formula : PI * r * r Where PI represents 3.14 Q.1(c) Explain difference between for, while and do while loop. [3] (A) For loop is used to repeat the same set of statements multiple times. Similarly while and do while loop. When the number of iterations are known then for loop is used. Otherwise while loop is preferable. [1 mark] e.g. display first n integers [1 mark] for (i = 0; i < n; i++) printf( %d, i); Using while loop Int i = 1; while(i <= n) Printf( %d, i); I++; 1

2 : F.E. SPA In do.. while, the condition is kept at the end and the loop will be executed even if the condition is false in the beginning. [1 mark] Q.1(d) Explain difference between call by value and call by reference. [4] (A) Earlier we have seen functions and function calls. When we passed the actual arguments, their copies are created into formal arguments; this method of calling function is called Call by Value. Let s take one example. int add (int a, int b) return (a+b); void main( ) int x = 2, y = 3, z; z = add(x, y); cout << Addition = << z << endl; Now here copies of x and y are created in a and b which are used for addition. If we change anything in a and b, then the changes are not reflected back in x and y. Hence call by value does not allow to change the actual argument. But here as we just want to add a and b and don t want to change them call by value suits the scenario. Hence in case of such situations where we don t want to change the formal arguments call by value is the best option. But now look at the following example. A function swap is written using call by value to swap the values of the integer variables. When the function is called by passing the reference / address of the variables then it is called as Call by reference. Let us see one example. void swap (int *a, int *b) int temp; temp = *a; *a = *b; *b= temp; void main ( ) int x,y; 2

3 May 13 Paper Solution cout<< Enter x and y\n ; cin >> x >> y ; swap (&x, &y) ; cout<< X = << x << Y= << y; Now suppose the user enters x and y as 2 and 3 respectively then the final output will be X = 3, Y = 2 which is not possible in the absence of call by reference. [4 marks for 4 points] Q.1(e) Explain any three string standard library functions. [3] (A) String Manipulating functions 1) Strlen (String Length) [1 mark] This functions is used to find the length of the given string. The functions takes string variable or string constant as a parameter and returns int value indicating the length. The null character is not considered while calculating length. e.g. int n = strlen ( STAR ); // n = 4 char n [ ] = TESTED OK ; int n = strlen (a); // n = 9 2) Strcmp (String Compares) [1 mark] The strcmp function compares two strings identified by the argument and has a value O. if they are equal. If they are not, it has the numeric difference between the first non-matching characters in the strings. strcmp (str1, str2) Where str and str 2 can either be string variables or constants. e.g. strcmp (str1, str2) strcmp (str1, STAR ); strcmp ( MAKE, MALE ); We are generally required to compare two strings lexicographically. i.e. alphabetically. If the answer is O. then two strings are equivalent. strcmp ( A, B ); Now mismatch occurs and hence it returns difference between Ascll value of A and B ie 1. Hence negative answer confirms that string 1 is alphabetically above than string 2. Hence we can draw the following conclusion. n = stcat (str1, str2); then if n = 0 str1 and str2 are same if n < 0 str1 is alphabetically above that str2 if n > 0 str2 is alphabetically above than str1 3

4 : F.E. SPA 3) Strcpy (String Copy) [1 mark] Strcpy basically serves the purpose of assigning one string to another. As we have seen that the direct assignment of string variable is not possible. Hence str1 = str2 is invalid. But this can be done as follows : strcpy (str1, str2); It copies str2 into str1. Hence original contents of str1 (if any) are lost but that of str2 remains same. Care should be taken that the length of array st1 should be atleast equal to array str2. In strcpy, str2 can be either string variable or string constant. But str1 has to be a string variable as it stores contents of str2 after copying. i.e. str2 gets assigned to strl. e.g. strcpy (str1, str2) str2 is copied int str1. strcpy (str1, MECH ); MECH is copied into str1. strcpy ( AUTO, RUN ); is invalid. Q.1(f) Explain reference and de-reference operators with example. [3] (A) Reference operator is used to compute address of a variable which can be stored using pointer. Dereference operator is used to retrieve back the value pointed by the pointer. [2 marks] e.g. int a = 20; [1 mark] int *p = &a; // reference int z = *p; // dereference Dereference operator is also called as indirection or value at operator. Q.2(a) Explain various storage classes used in c with example. [5] (A) Storage Classes : [for each point 1 mark, for example 1 mark] C++ provides total four storage classes as follow : 1) Automatic (auto) 2) External (extern) 3) Static (static) 4) Register (register) Automatic Variables (auto) The default storage class of a local variable is automatic. Hence it is actually a local variable. The scope is within the function in which it is declared and even the lifetime is till that function or a block is active. 4

5 May 13 Paper Solution Hence these variables are created on every call to that function and destroyed when the function is terminated. So they are not able to retain their value. int a; is equivalent to auto int a; The automatic variables store garbage values till they are explicitly initialized. External Variables (extern) The global variable s scope starts from the point of declaration till the end of the program. But the functions before the global variable declaration cannot access it. For Example; void main( ) p = p+5; // Compiler error : Undefined Symbol p int p; void function1 ( ) p = 10; Compiler doesn t identify p in main as it is declared after it. Hence to avoid this we can re-declare p as extern in main. Hence no extra memory space will be allocated to p but compiler understands that p is integer type of variable that is declared somewhere else. Hence it allows main to access it. void main( ) extern int p; ---- p = p + 5; int p; void function1 ( ) p = 10; 5

6 : F.E. SPA Static Variables (static) As we have seen regarding auto (local) variables that they get destroyed as soon as the function in which they are declared is terminated. Hence local variables cannot retain their values. The solution to this is static variables. These variables are local variables that are also defined within the function but their lifetime is till the end of the program. Hence they are created at the first call to the function but do not get destroyed when the function terminated. They also retain their values till the next time function is called. void main ( ) display( ); display( ); void display( ) static int a = 5; a = a + 1; cout << a; The output of the above program is 6 7 as the changed value of a in first call is carried till the second call. If a would not be made static the output would be 6 6. Register Variables (register) The variables which are frequently required should be kept in machine s registers rather than keeping it in the memory. So the access becomes rapid and a lot of time is conserved. This can be done by making the variables storage class as register. There are only few registers available, but C doesn t provide any restriction on making variable register although he internally converts register variable to non-register once the limit is reached. register int a; It makes a as register variable, the other characteristics it possesses like normal local variable 6

7 May 13 Paper Solution Q.2(b) Write a program to calculate sum of list by passing array to a [5] function. (A) Int findsum(int x[ ], int n) [5 marks] Int s=0,i; For (i=0; i < n; i++) s = s + x[i]; return s; Void main( ) int a[20], n, i; printf( Enter no of elements ); scanf( %d, &n); for (i = 0; i < n; i++) scanf( %d, &a[i]); int sum = findsum(a, n); printf( Sum = %d, sum); Q.2(c) Write a c-program to create an array of structure to store the [10] details of almost 100 employees. And sort it according to employee ID. Employee details are as follows: (i) Employee Name (ii) Employee ID (iii) Employee Salary (A) Typedef struct [10 marks] Char name[50]; Int id, sal; employee; Void sort(employee x[ ], int n) Int I, j; For (i = 0; i < n - 1; i++) For (j = 0; j < n - 1; j++) If (x[j].id > x[j+1].id) Employee t = x[j]; 7

8 : F.E. SPA X[j] = x[j+1]; X[j+1] = t; Void main( ) Employee e[100]; Int I; For (i = 0; i < 100; i++) scanf( %s%d%d,&e[i].name, &e[i].id,&e[i].sal); sort(e, n); For (i = 0; i < n; i++) printf( %s %d %d\n,e[i].name, e[i].id,e[i].sal); Q.3(a) Write an algorithm and draw flowchart to calculate roots of [6] quadratic equation. (A) Algorithm : [3 marks] 1) start 2) accept coefficients a, b, c 3) find d=b*b - 4ac 4) if d > 0 then 5) roots are real and disitinct 6) calculate root 1 and root 2 7) go to step 14 8) if d < 0 then 9) roots are imaginary 10) calculate real and imaginary part of root1 and root2 11) go to step 14 12) print roots are real and equal 13) calculate root 14) stop 8

9 May 13 Paper Solution Flowchart : [3 marks] Start accept coefficient a, b, c d = b * b 4 + a * c No yes d > 0 r 1 = b b 4ac r 2 = b b 4ac 2 2 display r 1, r 2 Stop Q.3(b) Write a program to display pascal triangle. [6] A A B A B C A B C D A B C D E (A) Void main( ) [6 marks] Int n, I, j; Char ch= A ; Printf( Enter no of lines ); 9

10 : F.E. SPA Scanf( %d, &n); For (i = 1; i <= n; i++) Ch= A ; for (j = 1; j <= I; j++) Printf( %c, ch++); Printf( \n ); Q.3(c) Explain recursion concept. Write a program to display Fibonacci [8] series using recursion. (A) Recursion is the process of calling a function within itself. Hence recursion tries to make a parallel option to the iterative procedures. As the definition suggests, each time the function is called it will call itself again within its body hence it feels that it will result in infinite iterations. But the recursion has to be controlled properly that is after a certain iterations the function should not allowed to call itself again. [4 marks] For any program to solve with the recursion the following three prerequisites can be listed out. 1) A program should be recursive in nature then only it becomes convenient to use a recursive approach. All iterative procedures may not be converted into recursive efficiently. 2) There should be a terminating condition which when becomes false, the recursion stops that is the same function is not called further. 3) Even the care has to be taken that during the execution of the program the above-mentioned terminating condition should be reached. Otherwise again the program will go into infinite loops. Now lets take some examples for a comparative study. In recursive we get two terminating condition as : fibo(1) = 1 and fibo (2) =1 For general value of n we can use the formula as, fibo (n) = fibo (n-1) + fibo (n-2) Again a recursion can be seen clearly and as we are decrementing value of n, the terminating conditions will surely be reached. Recursive Solution [4 marks] /*********** Recursive Fibo *********/ int fibo (int n) if (n==1 n==2) 10

11 return 1; return (fibo(n-1)+fibo(n-2)); void main() int num; clrscr( ); cout<<enter a number\n"; cin>>num; cout<<"required fibo number is <<fibo(num); getch( ); May 13 Paper Solution Q.4(a) Write an algorithm to sort set of numbers in ascending order. For [6] above problem in which cases time complexity is calculated. (A) Algorithm : [4 marks] 1) start 2) accept array of n integers 3) for i= 0 to n-1 do 4) for j=0; to n-1 do 5) if x(j) > x(j+1) then 6) exchange x(j) with x(j+1) 7) print elements of the array 8) stop Time complexity is calculated for two nested for loops. Outer for loop is executed for (n 1) times. Whereas inner for loop is executed for (n 1) times. The overall execution of for loop is tabulated below : Outer for loop Inner for loop i = 0 j = 0 j = 1 j = 2 j = 3 j = n 1 i = 1 j = 0 j = 1 j = n 1 [2 marks] 11

12 : F.E. SPA i = n 1 j = 0 j = 1 j = n 1 hence total number of for loop execution is (n 1) * (n 1) n 2 hence time complexity of above for loop is T(n 2 ). Q.4(b) Write a program to display Armstrong nos between 1 to [6] (A) Int ispalin(int num) [6 marks] int r = 0, u, k = num; While(num!=0) U = num%10; R = r*10 + u; Num = num%10; if (k==r) Return 1; Else return 0; Void main( ) int I; for (i=0; i<=1000;l i++) if (ispalin(i)==1) Pritnf( %d,i) Q.4(c) Write a program to calculate matrix multiplication and transpose [8] for a matrix. (A) #include <stdio.h> [8 marks] int main( ) int m, n, p, q, c, d, k, sum = 0; int first[10][10], second[10][10], multiply[10][10]; 12

13 May 13 Paper Solution printf("enter the number of rows and columns of first matrix\n"); scanf("%d%d", &m, &n); printf("enter the elements of first matrix\n"); for ( c = 0 ; c < m ; c++ ) for ( d = 0 ; d < n ; d++ ) scanf("%d", &first[c][d]); printf("enter the number of rows and columns of second matrix\n"); scanf("%d%d", &p, &q); if ( n!= p ) printf("matrices with entered orders can't be multiplied with each other.\n"); else printf("enter the elements of second matrix\n"); for ( c = 0 ; c < p ; c++ ) for ( d = 0 ; d < q ; d++ ) scanf("%d", &second[c][d]); for ( c = 0 ; c < m ; c++ ) for ( d = 0 ; d < q ; d++ ) for ( k = 0 ; k < p ; k++ ) sum = sum + first[c][k]*second[k][d]; multiply[c][d] = sum; sum = 0; printf("product of entered matrices:-\n"); for ( c = 0 ; c < m ; c++ ) for ( d = 0 ; d < q ; d++ ) printf("%d\t", multiply[c][d]); printf("\n"); return 0; 13

14 : F.E. SPA Q.5(a) Write a output for following program: [6] # include <stdio.h> Void main ( ) int x = 10, y, z; z = y = x; y = x; z = x ; x = x x ; printf( x = %d y = %d z = %d, x, y, z);p (A) Output [6 marks] Q.5(b) WAP to calculate sum of series 1/2 + 3/4 + 5/ n terms. [6] (A) Void main( ) [6 marks] int i; Float s=0; For (i=1; i<n; i=i+2) S= s+ (float) i / i+1; Printf( %f,s); Q.5(c) Write a program to calculate matrix multiplication and transpose [8] for a matrix. (A) #include<stdio.h> [8 marks] #include<conio.h> void main() int m, n, p, i, j, a[10][10], b[10][10] ; void MatMul (int a[10][10], int b[10][10], int m, int n, int p); clrscr(); printf("enter the number of rows and columns of matrix 1:"); scanf("%d %d", &m, &n); printf("enter the values of matrix l\n"); for(i=0;i<=m1;i++) for(j=0;j<=n1;j++) 14

15 printf("enter a value:"); scanf("%d", &a[i] [j]) ; printf("the number of rows for matrix 2 will be %d\n",n); printf("enter the columns of matrix 2:"); scanf("%d", &p); printf("enter the elements of matrix 2:\n"); for (i=0;i<=n1;i++) for(j=0;j<=p1;j++) printf("enter a value:"); scanf("%d", &b[i] [j]) ; MatMul (a, b, m, n, p); getch(); void MatMul (int a [10] [10], int b[10][10], int m, int n, int p) int i, j, k, c [10] [10] ; for (i=0;i<=m1; i++) for(j=0;j<=p1;j++) c[i] [j]=0; for(k=0;k<=n1;k++) c[i] [j]=c[i] [j]+a[i] [k]*b[k] [j]; printf("the resultant matrix is:\n"); for(i=0;i<=m1;i++) for(j=0;j<=p1;j++) printf("%d\t", c[i] [j]) ; May 13 Paper Solution 15

16 : F.E. SPA printf("\n"); #include<stdio.h> #include<conio.h> void main() int m, n, i, j, a [10] [10]; void transpose (int a[10][10], int m, int n); clrscr(); printf("enter the number of rows and columns:"); scanf("%d %d", &m, &n); for(i=0;i<=m1;i++) for(j=0;j<=n1;j++) printf("enter a value:"); scanf("%d", &a[i][j]); printf("the original Matrix is:\n"); for(i=0;i<=m-l;i++) for(j=0;j<=n1;j++) printf("%d \t", a[i] [j] ) ; printf("\n"); transpose(a, m, n); getch(); void transpose (int a[10] [10], int m, int n) int b[10] [10], i, j; for(i=0;i<=m1;i++) for(j=0;j<=n1;j++) b[j] [i]=a[i] [j] ; 16

17 printf("the transpose of this matrix is:\n"); for (i=0;i<=n1;i++) for(j=0;j<=m1;j++) printf("%d\t", b[i][j]); printf("\n"); May 13 Paper Solution Q.6(a) Explain difference between switch statement, ladder of if else and [6] nested if else. (A) The if statement by itself will execute a single statement, or a group of statements, when the expression following if evaluates to true. It does nothing when the expression evaluates to false. Can we execute one group of statements if the expression evaluates to true and another group of statements if the expression evaluates to false? Of course! This is what is the purpose of the else statement. For example :In a company an employee is paid as under If his basic salary is less than Rs. 1500, then HRA = 10% of basic salary and DA = 90% of basic salary. If his salary is either equal to or above Rs. 1500, then HRA = Rs. 500 and DA = 98% of basic salary. If the employee's salary is input through the keyboard write a program to find his gross salary. /* Calculation of gross salary */ main( ) floatbs, gs, da, hra ; printf ( "Enter basic salary " ) ; scanf ( "%f", &bs ) ; if ( bs< 1500 ) hra = bs * 10 / 100 ; da = bs * 90 / 100 ; else 17

18 : F.E. SPA hra = 500 ; da = bs * 98 / 100 ; gs = bs + hra + da ; printf ( "gross salary = Rs. %f", gs ) ; nested if-else. It is perfectly all right if we write an entire if-else construct within either the body of the if statement or the body of an else statement. This is called nesting of ifs. This is shown in the following program. /* A quick demo of nested if-else */ main( ) inti ; printf ( "Enter either 1 or 2 " ) ; if ( i == 1 ) printf ( "You would go to heaven!" ) ; else if ( i == 2 ) 18

19 printf ( "Hell was created with you in mind" ) ; else printf ( "How about mother earth!" ) ; May 13 Paper Solution LOOPS The versatility of the computer lies in its ability to perform a set of instructions repeatedly. This involves repeating some portion of the program either a specified number of times or until a particular condition is being satisfied. There are three methods by way of which we can repeat a part of a program. They are: (a) Using a for statement (b) Using a while statement (c) Using a do-while statement switch case statement A switch statement allows a variable to be tested for equality against a list of values. Each value is called a case, and the variable being switched on is checked for each switch case. Syntax: The syntax for a switch statement in C programming language is as follows: switch(expression) case constant-expression : statement(s); break; /* optional */ case constant-expression : statement(s); break; /* optional */ /* you can have any number of case statements */ default : /* Optional */ statement(s); The following rules apply to a switch statement: The expression used in a switch statement must have an integral or enumerated type, or be of a class type in which the class has a single conversion function to an integral or enumerated type. 19

20 : F.E. SPA You can have any number of case statements within a switch. Each case is followed by the value to be compared to and a colon. The constant-expression for a case must be the same data type as the variable in the switch, and it must be a constant or a literal. When the variable being switched on is equal to a case, the statements following that case will execute until a break statement is reached. When a break statement is reached, the switch terminates, and the flow of control jumps to the next line following the switch statement. Not every case needs to contain a break. If no break appears, the flow of control will fall through to subsequent cases until a break is reached. A switch statement can have an optional default case, which must appear at the end of the switch. The default case can be used for performing a task when none of the cases is true. No break is needed in the default case. Example: #include<stdio.h> int main () /* local variable definition */ char grade ='B'; 20

21 switch(grade) case'a': printf("excellent!\n"); break; case'b': case'c': printf("well done\n"); break; case'd': printf("you passed\n"); break; case'f': printf("better try again\n"); break; default: printf("invalid grade\n"); printf("your grade is %c\n", grade ); May 13 Paper Solution return0; When the above code is compiled and executed, it produces the following result: Well done Your grade is B [6 marks] Q.6(b) Write a recursive program to calculate factorial of accepted number. [6] (A) #include<stdio.h> [6 marks] int fact(int); int main( ) int num,f; printf("\nenter a number: "); scanf("%d", &num); f = fact(num); printf("\nfactorial of %d is: %d",num,f); return 0; 21

22 : F.E. SPA int fact(int n) if(n==1) return 1; else return(n*fact(n-1)); Q.6(c) Write a program to validate whether accepted string is palindrome [5] or not. (A) Implementation: [5 marks] #include<stdio.h> #include<conio.h> void main() int n=0,i; char a[100],rev[100]; clrscr(); printf("enter a string:"); gets(a); while(a[n]!='\0') n++; for(i=0;i<=(n-1);i++) rev[n-i-1]=a[i]; for(i=0;i<=n-1;i++) if(a[i]!=rev[i]) break; if(i==n) printf("the string is palindrome."); else printf("the string is not palindrome."); getch(); The above program checks whether source string is equal to target string (reverse of source string itself). 22

23 May 13 Paper Solution Q.6(d) Explain how to read the contents of file and write into the file with [3] Syntax. (A) Writing to a file [3 marks] #include <stdio.h> int main() int n; FILE *fptr; fptr=fopen("c:\\program.txt","w"); if(fptr==null) printf("error!"); exit(1); printf("enter n: "); scanf("%d",&n); fprintf(fptr,"%d",n); fclose(fptr); return 0; Reading from file #include <stdio.h> int main() int n; FILE *fptr; if ((fptr=fopen("c:\\program.txt","r"))==null) printf("error! opening file"); exit(1); /* Program exits if file pointer returns NULL. */ fscanf(fptr,"%d",&n); printf("value of n=%d",n); fclose(fptr); return 0; 23