Pointers MCQ Questions

[priyanka_20 (Priyanka Padiyar)]

I didn’t understood the working procedure for questions 6,7 and 8 like how to know what is the output of the given 3 programs

[aman212yadav (Aman Yadav)]

hello @priyanka_20
pls share the questions screenshot / text with me

[priyanka_20 (Priyanka Padiyar)]

Q­6 What will be the output of the following program?
#include
using namespace std;
int main() {
int arr[]={0,1,2,3,4};
int i, ptr;
for(ptr= arr, i=0; ptr+i <= arr+4; ptr++, i++) cout<<(ptr+i);
return 0;
}
A) 01234
B) 024
C) 234
D) 12340

7)Q­7 In the piece of code, arr[ ][ ] is a 2­D array and assume that the
contents of the 2­D array are already filled up. What is stored in the
variable sum at the end of the code segment?
int arr[3][3];
int i, sum=0; i = 0; while(i<3) {
sum += * ( * (arr+i)+(i++)); }
printf(“sum:%d”, sum);
A) Sum of all elements in the matrix
B) Sum of alternate elements in the matrix
C) Sum of the elements along the principal diagonal
D) None

Q­8 Would the following program compile?
#include
using namespace std;
int main() {
int a=10, *j; void *k; j=k=&a; j++;
k++; cout<<j<<k; return 0;
}Would the following program compile?
A) Yes
B) No

I didn’t understood the working procedure for the above programs

[aman212yadav (Aman Yadav)]

@priyanka_20

@priyanka_20

Q­6 What will be the output of the following program?
#include
using namespace std;
int main() {
int arr[]={0,1,2,3,4};
int i, ptr;
for(ptr= arr, i=0; ptr+i <= arr+4; ptr++, i++) cout<< (ptr+i);
return 0;
}
A) 01234
B) 024
C) 234
D) 12340

just follow the loop.
first it will print value at ptr which is 0. then they are incrementing ptr by 1 and i by 1
that means now it will print value ptr +1 + 1 ie 3rd element which is 2.
again it will incremnt by 2 and this time it will print ptr+4 ie 4th element which 4.
and loop will stop after that.
024 will be the answer.

@priyanka_20

7)Q­7 In the piece of code, arr[ ][ ] is a 2­D array and assume that the
contents of the 2­D array are already filled up. What is stored in the
variable sum at the end of the code segment?
int arr[3][3];
int i, sum=0; i = 0; while(i<3) {
sum += * ( * (arr+i)+(i++)); }
printf(“sum:%d”, sum);
A) Sum of all elements in the matrix
B) Sum of alternate elements in the matrix
C) Sum of the elements along the principal diagonal
D) None

to solve this problem u should know this->
a[i]= * (a+i)

similarly
a[i][j]= * (a[i]+j) = * ( * (a+i) + j )

so if u see this

@rakesh_anand

sum += * ( * (arr+i)+(i++))

this is nothing but arr[i][i++]

now i think u can calculate answer on ur own
correct answer will be-> Sum of the elements along the principal diagonal
because it compute sum of element at index (0,0 , 1,1 , 2,2 )

@priyanka_20

Q­8 Would the following program compile?
#include
using namespace std;
int main() {
int a=10, *j; void *k; j=k=&a; j++;
k++; cout<<j<<k; return 0;
}Would the following program compile?
A) Yes
B) No

Ans is No. it wont compile…
compile error is : error: invalid conversion from ‘void*’ to ‘int*’

why?
due to line j = k = &a.
in cpp, void* is a generic pointer and it can point to any type of element.
statements like k=&a, k=j or k=j=&a are all valid…
but j = k is not valid. if you assigns a generic type(void*) to a specific one(int*), you need to cast it. i.e. int* can not point to void*(since void* is superset)

so cast it… j = (int*)k will work.