Commvault Aptitude Test Papers
11 / 40
char *buff, *data;
void main ()
{
const char* const cp = buff;
cp = data; // Line 5
*cp='a'; // Line 6
}
ANo compilation error
BOnly Line 5
COnly Line 6
DBoth Line 5 and Line 6
Answer: Option D
Explanation:Here is no explanation for this answer
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12 / 40
#include <iostream>
using namespace std;
#define SQ(a) a*a
inline int square(int a)
{
return a*a;
}
int main () {
int i, j, k, l;
i = SQ(2 + 3);
j = square(2 + 3);
k = SQ(6 - 1);
l = square(6 - 1);
cout << i << " "<< j << " "<< k << " " << l;
return 0;
}
A11 25 -1 25
B11 11 -1 -1
C25 25 25 25
DNone of the above
Answer: Option A
Explanation:Here is no explanation for this answer
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13 / 40
Which statement is wrong in C++?
AA pointer to a constant can be used to view but not modify its target
BA canst pointer can't be modified but its target can
CThe inline specifier instructs the compiler to replace function calls with the code of the function body
Dconstant pointers to constants can be used to view but not modify its target
Answer: Option C
Explanation:Here is no explanation for this answer
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14 / 40
Which of the following programming technique/structures are not good for demand paged environment?
AStack
BSequential search
CBinary Search
DVector operations
Answer: Option C
Explanation:Here is no explanation for this answer
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15 / 40
Which of the following statement is incorrect?
AVolatile storage refers to main and cache memory and is usually fast.
BVolatile storage can survive system crashes or system power down.
CDisk and tapes (non volatile storage) can survive system crashes or system power down.
DStable storage (disk) refers to storage that technically can never be lost as there are redundant backup copies of data.
Answer: Option B
Explanation:Here is no explanation for this answer
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16 / 40
Consider the following program which comprises three source files main.c, fl.c, f2.c? What is the output of the program?
#include <stdio.h>
//test.c
void f1 ();
void f2();
static int a;
static int b;
void main()
{
f1 ();
f2();
printf("f5 = %d f6 = %d ", a, b );
}
//f2.c
int a;
int b;
void f2(){
printf("f3 = %d f4 = %d ", a,b);
}
//f1 .c
extern int a;
extern int b;
void f1 ()
{
b = 6;
a= 5;
printf("f1 = %d f2 = %d ", a,b);
}
Afl = 5 f2 = 6 f3 = 5 f4 = 6 f5 = 0 f6 = 0
Bfl = 5 f2 = 6 f3 = 5 f4 = 6 f5 = 5 f6 = 6
Cfl = 0 f2 = 0 f3 = 5 f4 = 6 f5 = 0 f6 = 0
Dfl = 5 f2 = 6 f3 = 0 f4 = 0 f5 = 5 f6 = 6
Answer: Option A
Explanation:Here is no explanation for this answer
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17 / 40
What are the number of nodes of left and right sub-tree of the binary tree if the data is inserted in the following order:\r\n 45, 15, 8, 5 6, 5, 65, 47, 12, 18, 10, 73, 50, 16, 61
A7 6
B6 7
C8 5
D5 8
Answer: Option A
Explanation:Here is no explanation for this answer
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18 / 40
Match the following with reference to the following program code.
#include
void main ()
{
int x [3][4] = {{1,6,9,12},{11,17,3,2},{20,23,4,5}};
int *n = &x;
}
(i) *(*(x + 1) + 1) (1) 9
(ii) *(*x + 1) + 3 (2) 13
(iii) *(n + 3) + 1 (3) 4
(iv) ++(*n++) + *n (4) 17
Ai - 3, ii - 1, iii - 2, iv - 4
Bi - 2, ii - 4, iii - 1, iv - 3
Ci - 4, ii - 3, iii - 2, iv - 1
Di - 4, ii - 3, iii - 1, iv - 2
Answer: Option C
Explanation:Here is no explanation for this answer
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19 / 40
Consider the following code segment in C to traverse a binary tree using the preorder
typedef struct tree {
int info;
struct *left;
struct *right;
}node;
void preorder(node *tree)
{
if (t)
{
Statementl
Statement2
Statement3
}
}
The above Statements should be
Apreorder(tree->right); preorder(tree->left); printf("%d", tree->info);
Bpreorder(tree->left); preorder(tree->right); printf("%d", tree->info);
Cpreorder(tree->left); printf("%d", tree->info); preorder(tree->right);
Dprintf("%d", tree->info); preorder(tree->left); preorder(tree->right);
Answer: Option D
Explanation:Here is no explanation for this answer
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20 / 40
Which of the following are true with respect to deadlock conditions?
1) Mutual exclusion: a resource that cannot be used by more than one process or thread at a time
2) Hold and wait: processes already holding resources may request new resources
3) No preemption: No resource can be forcibly removed from a process holding it, resources can be released only by the explicit action of the process
4) Circular wait: two or more processes form a circular chain where each process waits for a resource that the next process in the chain holds
AOnly 1
BOnly 1 and 2
COnly l, 2 and 3
DAll the 4 conditions
Answer: Option D
Explanation:Here is no explanation for this answer
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Here is the list of questions asked in Aptitude Test for Commvault Commvault Aptitude Test Papers. Practice Commvault Written Test Papers with Solutions and take Q4Interview Commvault Online Test Questions to crack Commvault written round test. Overall the level of the Commvault Online Assessment Test is moderate. Only those candidates who clear the written exam will qualify for the next round, so practic all the questions here and take all the free tests before going for final selection process of Commvault