Arrays Questions and Answers

Q: What is printed by the following ANSI C program?#include\(\lt\)stdio.h\(\gt\)int main(int argc, char *argv[]){ int a[3][3][3] = {{1, 2, 3, 4, 5, 6, 7, 8, 9}, {10, 11, 12, 13, 14, 15, 16, 17, 18}, {19, 20, 21, 22, 23, 24, 25, 26, 27}}; int i = 0, j = 0, k = 0; for( i = 0; i \(\lt\) 3; i++ ){ for(k = 0; k \(\lt\) 3; k++ ) printf("%d ", a[i][j][k]); printf("\n"); } return 0;}

The program prints 1 2 3 / 10 11 12 / 19 20 21, one group per line.Each inner brace list initialises one 3 by 3 block a[i] with nine values, filled row by row, so a[0] holds 1 to 9, a[1] holds 10 to 18 and a[2] holds 19 to 27. In particular a[i][0] is the first row of each block: 1 2 3, then 10 11 12, then 19 20 21.The loops vary i and k but never change j, which stays 0. So for each i the program prints a[i][0][0], a[i][0][1], a[i][0][2] and then a newline, which is the first row of each block.'1 4 7 ...' would come from varying j instead of k. '1 2 3 / 4 5 6 / 7 8 9' prints the three rows of a[0] only, as if i stayed 0 while j changed. '1 2 3 / 13 14 15 / 25 26 27' moves along the diagonal blocks, as if j rose with i.

Q: What will be the output of the program ? #include <stdio.h> int main() { int arr[1]={10}; printf("%d\n", 0[arr]); return 0; }

Correct answer: 10Subscripting is defined as a[i] being exactly *(a + i). Addition commutes, so *(a + i) is also *(i + a), which is i[a]. Therefore 0[arr] means arr[0], printing 10. It is legal but nobody writes it outside puzzles.

Q: What is meaning of following declaration ? int arr[20];

Correct answer: Integer Array of size 20int arr[20]; reserves room for 20 values of type int, laid out consecutively, with subscripts 0 to 19. Read the declaration outwards from the name: arr is an array of 20, of int.

Q: What is the value of a[4]? int a[5]={1,2,4,1,0}

Correct answer: 0The initialiser list fills the array in order, so a[0] = 1, a[1] = 2, a[2] = 4, a[3] = 1 and a[4] = 0 - the last value in the list, since indexing starts at 0.

Q: Choose the correct statements

Correct answer: Both A and BA is true: every element of an array has the same type and the same storage class, which is what makes uniform address arithmetic possible. B is true: the number of subscripts is the dimension - a[3][4] is two-dimensional. The remaining option is false, since a subscript may perfectly well be zero; indeed index 0 is the first element.

Q: A file is preferable to an array of structures because

Correct answer: All of the aboveAll three reasons hold. A file persists after the program that wrote it ends, while an array vanishes with the process. A file need not be sized in advance, so no space is reserved and wasted. And the operating system provides tools - copy, search, sort, back up - that work on files and not on a program’s memory.

Q: The information about an array used in program will be stored in

Correct answer: Dope VectorA dope vector is the descriptor the implementation keeps for an array: its base address, element size, number of dimensions and the bounds of each. It is what makes address computation and bounds checking possible. The symbol table holds compile-time names and types; an activation record holds a call’s local storage.

Q: What is the syntax for accessing the element at index 7 of an array a?

Correct answer: a[7]C uses square brackets for subscripting, so the element at index 7 is a[7]. Braces are for initialiser lists, parentheses would be read as a function call, and quotes make a string literal. Note that a[7] is the eighth element, since indexing starts at 0.

Q: If x is an array of integers, then the value of &x[i] is same as

Correct answer: x+iBy definition x[i] is *(x + i), so taking the address of both sides gives &x[i] == x + i. The scaling is already built into pointer arithmetic: adding 1 to an int * advances by sizeof(int) bytes. That is why the two options that multiply or add sizeof(int) explicitly are wrong - they would scale twice.

Q: A one dimensional array A has indices 1....75. Each element is a string and takes up three memory words. The array is stored at location 1120 decimal. The starting address of A[49] is

Correct answer: 1264Indices start at 1 here, not 0, so \(A[49]\) is the 49th element and 48 elements precede it. Each takes 3 words:\[ 1120 + 3 \times (49 - 1) = 1120 + 144 = 1264 \]Forgetting the 1-based start gives 1267, which is the distractor.

What is printed by the following ANSI C program?

#include\(\lt\)stdio.h\(\gt\)
int main(int argc, char *argv[])
{
    int a[3][3][3] =
    {{1, 2, 3, 4, 5, 6, 7, 8, 9},
     {10, 11, 12, 13, 14, 15, 16, 17, 18},
     {19, 20, 21, 22, 23, 24, 25, 26, 27}};
  int i = 0, j = 0, k = 0;
  for( i = 0; i \(\lt\) 3; i++ ){
    for(k = 0; k \(\lt\) 3; k++ )
      printf("%d ", a[i][j][k]);
    printf("\n");
  }
  return 0;
}

A.1 2 3 / 10 11 12 / 19 20 21
B.1 4 7 / 10 13 16 / 19 22 25
C.1 2 3 / 4 5 6 / 7 8 9
D.1 2 3 / 13 14 15 / 25 26 27

Exam Year(s)

GATE CS - 2022,

Answer: A

The program prints 1 2 3 / 10 11 12 / 19 20 21, one group per line.

Each inner brace list initialises one 3 by 3 block a[i] with nine values, filled row by row, so a[0] holds 1 to 9, a[1] holds 10 to 18 and a[2] holds 19 to 27. In particular a[i][0] is the first row of each block: 1 2 3, then 10 11 12, then 19 20 21.

The loops vary i and k but never change j, which stays 0. So for each i the program prints a[i][0][0], a[i][0][1], a[i][0][2] and then a newline, which is the first row of each block.

'1 4 7 ...' would come from varying j instead of k. '1 2 3 / 4 5 6 / 7 8 9' prints the three rows of a[0] only, as if i stayed 0 while j changed. '1 2 3 / 13 14 15 / 25 26 27' moves along the diagonal blocks, as if j rose with i.

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What will be the output of the program ?

#include <stdio.h>

int main()
{
    int arr[1]={10};
    printf("%d\n", 0[arr]);
    return 0;
}

A.1
B.10
C.6

Answer: B

Correct answer: 10

Subscripting is defined as a[i] being exactly *(a + i). Addition commutes, so *(a + i) is also *(i + a), which is i[a]. Therefore 0[arr] means arr[0], printing 10. It is legal but nobody writes it outside puzzles.

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What is meaning of following declaration ?

int arr[20];

A.Integer Array of size 20
B.None of these
C.Array of size 20 that can have integer address
D.Array of Size 20

Answer: A

Correct answer: Integer Array of size 20

int arr[20]; reserves room for 20 values of type int, laid out consecutively, with subscripts 0 to 19. Read the declaration outwards from the name: arr is an array of 20, of int.

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What is the value of a[4]?

int a[5]={1,2,4,1,0}

A.1
B.10
C.0
D.Error

Answer: C

Correct answer: 0

The initialiser list fills the array in order, so a[0] = 1, a[1] = 2, a[2] = 4, a[3] = 1 and a[4] = 0 - the last value in the list, since indexing starts at 0.

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Choose the correct statements

A.All The elements of the array should be of the same data type and storage class
B.The number of subscripts determines the dimension of the array
C.Both A and B
D.In an array definition, the subscript can be any expression yielding a non-zero integer value

Answer: B

Correct answer: Both A and B

A is true: every element of an array has the same type and the same storage class, which is what makes uniform address arithmetic possible. B is true: the number of subscripts is the dimension - a[3][4] is two-dimensional. The remaining option is false, since a subscript may perfectly well be zero; indeed index 0 is the first element.

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A file is preferable to an array of structures because

A.file lives even after the program that created it terminates
B.memory space will not be wasted
C.there are many system tools to manipulate files
D.All of the above

Answer: D

Correct answer: All of the above

All three reasons hold. A file persists after the program that wrote it ends, while an array vanishes with the process. A file need not be sized in advance, so no space is reserved and wasted. And the operating system provides tools - copy, search, sort, back up - that work on files and not on a program’s memory.

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The information about an array used in program will be stored in

A.Symbol Table
B.Activation Record
C.Dope Vector
D.Both A and B

Answer: C

Correct answer: Dope Vector

A dope vector is the descriptor the implementation keeps for an array: its base address, element size, number of dimensions and the bounds of each. It is what makes address computation and bounds checking possible. The symbol table holds compile-time names and types; an activation record holds a call’s local storage.

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What is the syntax for accessing the element at index 7 of an array a?

A.a{7}
B.a(7)
C.a"7"
D.a[7]

Answer: D

Correct answer: a[7]

C uses square brackets for subscripting, so the element at index 7 is a[7]. Braces are for initialiser lists, parentheses would be read as a function call, and quotes make a string literal. Note that a[7] is the eighth element, since indexing starts at 0.

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If x is an array of integers, then the value of &x[i] is same as

A.&x[i-1] + sizeof (int)
B.x + sizeof (int) * i
C.x+i
D.none of these

Answer: A

Correct answer: x+i

By definition x[i] is *(x + i), so taking the address of both sides gives &x[i] == x + i. The scaling is already built into pointer arithmetic: adding 1 to an int * advances by sizeof(int) bytes. That is why the two options that multiply or add sizeof(int) explicitly are wrong - they would scale twice.

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10.

A one dimensional array A has indices 1....75. Each element is a string and takes up three memory words. The array is stored at location 1120 decimal. The starting address of A[49] is

A.1264
B.1164
C.1167
D.1267

Answer: A

Correct answer: 1264

Indices start at 1 here, not 0, so \(A[49]\) is the 49th element and 48 elements precede it. Each takes 3 words:

\[ 1120 + 3 \times (49 - 1) = 1120 + 144 = 1264 \]

Forgetting the 1-based start gives 1267, which is the distractor.

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