Arrays Questions and Answers

Answer: C

Correct answer: The program may crash if some important data gets overwritten.

C does no bounds checking. Writing past the end of an array simply stores to whatever address the index computes to, so the effect depends on what happens to be there: often nothing visible, sometimes a corrupted variable, sometimes a crash when a return address or a pointer is overwritten. The compiler cannot generally report it, the element is not zeroed, and the array certainly does not grow.

Answer: C

Correct answer: Both A and B

A string parameter may be written either as char s[] or as char *s, and the two are exactly equivalent in a parameter list: an array argument decays to a pointer to its first element, so the function receives an address either way. Outside a parameter list the two forms are quite different.

Answer: C

Correct answer: 1, 1, 1
2, 2, 2
3, 3, 3
3, 4, 4

p holds the five addresses arr … arr+4, and ptr walks along p. Take the four statements in turn.

• ptr++ - ptr = p+1. Then ptr-p is 1, *ptr-arr is 1, **ptr is arr[1] = 1 → 1, 1, 1
• *ptr++ - the ++ binds to ptr, and the dereference is thrown away, so ptr = p+2 → 2, 2, 2
• *++ptr - again only ptr moves, to p+3 → 3, 3, 3
• ++*ptr - here the increment applies to *ptr, that is to p[3], which becomes arr+4. ptr itself has not moved → ptr-p = 3, *ptr-arr = 4, **ptr = arr[4] = 4 → 3, 4, 4

The point of the question is that the last line increments what the pointer points to, while the three before it increment the pointer.

Answer: B

Correct answer: 2-D Array

int RollNum[30][4]; has two subscripts, so it is a two-dimensional array - 30 rows of 4 columns, 120 integers in all. The number of dimensions is simply the number of bracket pairs, not the values inside them.

Answer: A

Correct answer: int geeks[20];

A C array declaration is type, name, then the size in square brackets: int geeks[20];. int geeks; declares a single integer, braces are not used for the dimension, and array is not a type keyword in C.

Answer: B

Correct answer: var a:array[REAL]of REAL;

An array index type must be an ordinal type - one whose values can be counted off, so that each has a definite position. Enumerations, characters and booleans all qualify, which is why the other three definitions are legal. REAL is not ordinal: there is no “next” real number, so it cannot index an array.

Answer: B

Correct answer: Compilation Error

A structure declaration describes a type, not an object, so its members cannot be given initialisers - char name[] = "GeeksforGeeks"; and int no_of_pages = 200; are both rejected. A member array also cannot have an unspecified size. The code therefore fails to compile, and the uninitialised ptr never becomes an issue.

Answer: A

Correct answer: sixth row

arr is an array of rows, so arr[5] is the row at index 5 - the sixth row, since indexing starts at 0. It has type int[12] and decays to a pointer to that row’s first element. The column count never enters into which row is selected.

Answer: B

Correct answer: n + ⌈log₂n⌉ − 2

Run a knockout tournament. Finding the largest takes \(n - 1\) comparisons, and the second largest must be one of the elements that lost directly to the winner - there are \(\lceil \log_2 n \rceil\) of them, since that is the depth of the tournament. Finding the best of those costs \(\lceil \log_2 n \rceil - 1\) more:

The naive approach - two separate passes - would cost about \(2n\), so the tournament saves nearly half.

Answer: A

Correct answer: Compile time

An ordinary array such as int a[20]; has its size fixed in the source, so the space is reserved when the program is compiled and laid out - static, compile-time allocation. Run-time allocation is what malloc and new do, where the size can depend on a value computed while the program runs.