Test: Arrays - 2 - Back-End Programming MCQ

There is compilation error in the declaration " int a[][] = {{1,2},{3,4}};". Except the first dimension, every other dimension must be specified.
int arr[] = {5, 6, 7, 8} //valid
int arr[][5] = {}; //valid
int arr[][] = {}; //invalid
int arr[][10][5] = {}; //valid
int arr[][][5] = {}; //invalid

The final expression can be simplified in form ofi, j and k by following the intermediate code steps in reverse order
t5 = X[t4]
   = X[t3 + t2]
   = X[t1 + t0 + t2]
   = X[i*1024 + j*32 + k*4]
   = X + i*1024 + j*32 + k*4 
Since k is multiplied by 4, the array must be an int array. We are left with 2 choices (A and B) among the 4 given choices. X[i][j][k]'th element in one dimensional array is equivalent to X[i*M*L + j*L + k]'th element in one dimensional array (Note that multi-dimensional arrays are stored in row major order in C). So we get following equations
j*L*4 = j*32, we get L = 8 (4 is the sizeof(int))
i*1024 = i*M*L*4, we get M = 1024/32 = 32
Therefore option A is the only correct option as M and L are 32 and 8 respectively only in option A.

We are just printing the values of first 3 values.

Here arrPtr is a global array of pointers to int. It should be noted that global variables such arrPtr are initialized to ZERO. That’s why all are elements of arrPtr are initialized implicitly to ZERO i.e. correct answer is b.

There’s no issue with definition of arr1 and arr2. In definition of these arrays, the mention of array size using variable is ok as per C standard but these types of arrays can’t be initialized at the time of definition. That’s why initialization of arr1 is incorrect. But initialization of arr2 is done correctly. Right answer is C.

If you look for loop carefully, you will notice that it assigns sum variable to some value in if condition and increments it in the else condition. On further thought, it would be clear that this loop stores sum of increasing subsequence of positive integers in sum variable and max of sum in maxsum. Hence, maxsum - maximum sum of increasing subsequence of positive integers will get printed out when this program is executed, which is 3 + 4 = 7. This solution is contributed by Vineet Purswani //output will be 3+4 =7 {for || if 1st argument is true 2nd argument will not be calculated, and if 1st argument is false, 2nd argument will be calculated}

Another Solution : When i=1 -> i==0 is false, but a[i]<0 is true so condition (1) is true.Now if (sum > maxsum) is true, since sum=2 and maxsum=0.So maxsum=2. sum = (a [i] > 0) ? a [i] : 0; , sum=0 since a[i]<0.
When i=2 -> i==0 is false, a[i]<0 is true and so condition (1) is true.Now if (sum > maxsum) is false, since sum=0 and maxsum=2.Since sum = (a [i] > 0) ? a [i] : 0; , sum=0 since a[i]<0.
When i=3 -> i==0 is false , a[i]<0 is false and a [i] < a [i – 1] is false so condition (1) is false. Now sum += a [i] = 3.
When i=4 -> i==0 is false , a[i]<0 is false and a [i] < a [i – 1] is false so condition (1) is false. Now sum += a [i] = 7.
When i=5 -> i==0 is false , a[i]<0 is false and a [i] < a [i – 1] is true so condition (1) is true. sum > maxsum is true, since sum=7 and maxsum=2,so maxsum=7.Since sum = (a [i] > 0) ? a [i] : 0, so sum=2 since a[5]>0.