Arrays And Strings - Study Mode
[#81] The . . . . . . . . function appends not more than n characters.
Correct Answer
(C) strncat()
[#82] What will be the output of the following code? void main()
{
int a[10]
printf("%d %d", a[-1], a[12])
}
Correct Answer
(D) Garbage vlaue Garbage Value
Explanation
Solution: In c compiler does not check array with its bounds, value at the computed location is displayed.
[#83] What does the following declaration mean? int ( *ptr ) [ 10 ]
Correct Answer
(B) ptr is a pointer to an array of 10 integers
Explanation
Solution: The declaration int (*ptr)[10]
means that ptr is a pointer to an array of 10 integers. Here's the breakdown: int specifies the type of elements in the array (integers). (*ptr) indicates that ptr is a pointer. [10] specifies that the pointer points to an array of 10 integers. Thus, ptr is not an array of pointers, nor an array of integers, but a pointer to an array of integers with a size of 10.
[#84] What will be the output of the program if the array begins at 65472 and each integer occupies 2 bytes? #include<stdio.h>
void main()
{
int a[3][4] = {1, 2, 3, 4, 4, 3, 2, 1, 7, 8, 9, 0}
printf("%u, %u", a+1, &a+1)
}
Correct Answer
(C) 65480, 65496
Explanation
Solution: >> int a[3][4] = {1, 2, 3, 4, 4, 3, 2, 1, 7, 8, 9, 0}
The array a[3][4] is declared as an integer array having the 3 rows and 4 colums dimensions. >> printf ("%u, %u
", a+1, &a+1)
The base address(also the address of the first element) of array is 65472. For a two-dimensional array like a reference to array has type "pointer to array of 4 ints". Therefore, a+1 is pointing to the memory location of first element of the second row in array a. Hence 65472 + (4 ints * 2 bytes) = 65480 Then, &a has type "pointer to array of 3 arrays of 4 ints", totally 12 ints. Therefore, &a+1 denotes "12 ints * 2 bytes * 1 = 24 bytes". Hence, begining address 65472 + 24 = 65496. So, &a+1 = 65496 Hence the output of
[#85] What will be the output of the program? #include<stdio.h>
int main()
{
int arr[1] = {10}
printf("%d", 0[arr])
return 0
}
Correct Answer
(C) 10
Explanation
Solution: >> int arr[1]={10}
The variable arr[1] is declared as an integer array with size '2' i.e. arr[0] and arr[1] and it's first element is initialized to value '10'(means arr[0]=10) and arr[1] = garbage value or zero >> printf ("%d", 0[arr])
It prints the first element value of the variable arr. Hence the output of the program is 10.