B. C. A. ( Semester - II ),2013 Question Paper,‘C’ PROGRAMMING,University Of Pune Question Paper

University Of Pune Question Paper
B. C. A. ( Semester - II ) Examination - 2013
‘C’ PROGRAMMING
(New 2008 Pattern)
Time : 3 Hours] [Max. Marks : 80
Q.1) Answer the following : (Any Ten) [10x2=20]
(1) Define Pointer to Pointer. Give example.
(2) What are the limitations of Array ?
(3) List the ways in which an array is passed to function.
(4) Give the syntax of Declaring Union.
(5) Apply Left Shift Operator on the data 0000000011010111 and
Shift Data by 5 bits.
(6) Give syntax and use of ftell( ).
(7) List the different modes in which file is opened.
(8) What is the significance of argv[ ] ?
(9) Give the syntax of File Inclusion Directive.
(10) Define String. Which character is used to terminate a string ?
(11) List the different types of File.
(12) How are structure elements accessed ?
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Seat
No.
Q.2) Answer the following : (Any Four) [5x4=20]
(a) What is Dynamic Memory Allocation ? Explain any two
functions of Dynamic Memory Allocation.
(b) Define Structure. What do you mean by Nested Structure ?
Explain with example.
(c) Define File. Explain different operations performed on File.
(d) Compare Structure and Union with example.
(e) What is an Array ? How to represent two dimensional arrays
in Memory ?
Q.3) Attempt the following : (Any Four) [5x4=20]
(a) Write a ‘C’ program to write a macro definition for checking
whether a character is alphabet or not.
(b) Write a ‘C’ program to store all prime numbers in an array and
display this array.
(c) Write a ‘C’ program to create a structure containing Student Roll
No., Name and Marks. Display student information having marks
greater than 60.
(d) Write a ‘C’ program to display alternate character in existing
file.
(e) Write a ‘C’ program to accept n numbers from user. Sort elements
in ascending order using Dynamic Memory Allocation.
Q.4) Trace the output and justify : (Any Four) [5x4=20]
(a) main( )
{
int c[] = {2, 8, 3, 4, 4, 6, 7, 5};
int j, *p = c, *q = c;
for (j = 0; j < 5; j++)
[4373]-203 2 Contd.
{
printf(“%d”, *c);
++ q;
}
for (j = 0, j < 5; j++)
{
printf(“%d”, * p);
++ p;
}
}
(b) #define SQR(x) (x * x)
int main( )
{
int a, b = 3;
a = SQR (b + 2);
printf(“%d\n”, a);
return 0;
}
(c) void main( )
{
printf(“\n %d %d %d”, size of(‘3’), sizeof(“3”), size of(3));
}
(d) main( )
{
char * a[5] = {“GOOD”, “BAD”, “UGLY”, “WICKED”, “NICE”};
printf(“%s\n”, a[0]);
printf(“%s\n”, * (a + 2));
printf(“%c\n”, * (a[2] + 2));
printf(“%s\n”, a[3]);
printf(“%c\n”, * (a[3] + 2));
}
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(e) main( )
{
int a[5] = {5, 1, 15, 20, 25};
int i, j, m;
i : ++ a[1];
j = a[1] ++;
m = a[i++];
printf(“%d %d %d”, i, j, m);

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