# Ds Lab Quiz Cse B

10 Questions | Total Attempts: 37  Settings  • 1.
The postfix form of the expression (A+ B)*((C*D− E)*F / G) is
• A.

AB+ CD*E − FG /**

• B.

AB + CD* E − F **G /

• C.

AB + CD* E − *F *G /

• D.

AB + CDE * − * F *G / AB + CDE * − * F *G / AB + CDE * − * F *G / AB + CDE * − * F *G /

• 2.
Overflow condition for simple linear queue array implementation is:
• A.

Front = Maximum -1

• B.

Rear = Maximum -1

• C.

Front = Rear

• D.

Front = -1

• 3.
Memory arrangement in C language is:
• A.

Row Major

• B.

Column Major

• C.

Diagonal

• D.

None

• 4.
What is the following code segment doing?void fn( ){               char c;               cin.get(c);               If (c != ‘\0’) {       fn( );      cout.put(c);  }}
• A.

The string entered is printed as it is.

• B.

The string entered is printed in reverse order.

• C.

It will go in an infinite loop.

• D.

It will print an empty line.

• 5.
When determining the efficiency of algorithm, the space factor is measured by
• A.

Counting the maximum memory needed by the algorithm

• B.

Counting the minimum memory needed by the algorithm

• C.

Counting the average memory needed by the algorithm

• D.

Counting the maximum disk space needed by the algorithm

• 6.
Struct node {      int data;      struct node *next;}*start = NULL;Consider the above representation and predict what will be printed on the screen by following statement ?start-->next-->data
• A.

None of these

• B.

Access the “data” field of 2nd node

• C.

Access the “data” field of 3rd node

• D.

Access the “data” field of 1st node

• 7.
Which one of the following represents an overflow condition for circular queue:
• A.

Front=0, Rear=Max-1

• B.

Front=Rear

• C.

Rear=0. Front=0

• D.

Front=-1, Rear=-1

• 8.
Random access is possible in:
• A.

• B.

Array

• C.

Queue

• D.

Stack

• 9.
Consider the following code:struct node{    int data;    struct node* next;}; static void reverse(struct node** head_ref){    struct node* prev   = NULL;    struct node* current = *head_ref;    struct node* next;    while (current != NULL)    {next  = current->next;         current->next = prev;          prev = current;        current = next;    }    *head_ref = prev;} The above code:
• A.

• B.

Exchange Values of consecutive nodes

• C.

• D.

Assigns node value to each other in random order

• 10.
/* Assume that n is greater than or equal to 0 */void fun2(int n){  if(n == 0)    return;   fun2(n/2);  printf("%d", n%2);} The function will display:
• A.

Counting from 1 to n

• B.

Reverse Counting from n to 1

• C.

Binary Equivalent of number n

• D.

N is recursively divided by 2 Back to top