In general, the index of the first element in an array is __________
Answer: D
Correct answer: 0
Array indexing begins at 0 in C, C++, Java, Python and most other languages. The reason is that the index is an offset from the base address: the address of element \(i\) is \(\text{base} + i \times \text{size}\), so the first element sits at an offset of zero.
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What will happen if in a C program you assign a value to an array element whose subscript exceeds the size of array?
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.
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A string that is a formal parameter can be declared
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.
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What will be the output of the program ?
#include <stdio.h>
int main()
{
static int arr[] = {0, 1, 2, 3, 4};
int *p[] = {arr, arr+1, arr+2, arr+3, arr+4};
int **ptr=p;
ptr++;
printf("%d, %d, %d\n", ptr-p, *ptr-arr, **ptr);
*ptr++;
printf("%d, %d, %d\n", ptr-p, *ptr-arr, **ptr);
*++ptr;
printf("%d, %d, %d\n", ptr-p, *ptr-arr, **ptr);
++*ptr;
printf("%d, %d, %d\n", ptr-p, *ptr-arr, **ptr);
return 0;
}
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, 4The point of the question is that the last line increments what the pointer points to, while the three before it increment the pointer.
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Below is an example of -
int RollNum[30][4];
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.
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Which of the following correctly declares an array?
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.
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Which of the following is an illegal array definition?
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.
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Predict the output of below code:
#include <stdio.h>
int main()
{
struct site
{
char name[] = "GeeksforGeeks";
int no_of_pages = 200;
};
struct site *ptr;
printf("%d",ptr->no_of_pages);
printf("%s",ptr->name);
getchar();
return 0;
}
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.
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If arr is a two dimensional array of 10 rows and 12 columns, then arr[5] logically points to the
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.
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Minimum number of comparisons required to compute the largest and second largest element in array is
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:
\[ (n - 1) + \lceil \log_2 n \rceil - 1 = n + \lceil \log_2 n \rceil - 2 \]The naive approach - two separate passes - would cost about \(2n\), so the tournament saves nearly half.
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