Question

Using a loop, write code that rotates the members of a 32-bit integer array forward one position. array dword 11,12,13,14,15,16,17,18,19,20 The value at the end of the array must wrap around to the first position. For example, the array [11,12,13,14] would be transformed into [14,11,12,13] Print the content of the array before and after shifting in the console window.

Answer 1

To rotate a 32-bit integer array forward by one position, store the last element, shift each element forward using a loop, and set the first element to the previously stored value. This manipulation results in the desired rotation, which can be printed to the console.

Step-by-step explanation:

To rotate the members of a 32-bit integer array forward by one position, you can write a loop in a programming language such as C, C++, or Java. Here is an example using pseudocode that can be easily translated into one of these programming languages:

int array[] = {11, 12, 13, 14, 15, 16, 17, 18, 19, 20};
int size = sizeof(array) / sizeof(array[0]);
int lastElement = array[size - 1];

for (int i = size - 1; i > 0; i--) {
array[i] = array[i - 1];
}
array[0] = lastElement;

// Print the array before and after shifting
for (int i = 0; i < size; i++) {
printf("%d ", array[i]);
}

This code assigns the last array element to a temporary variable, shifts each element one position forward in a loop, and finally assigns the stored last element to the first position of the array. This effectively rotates the array items forward. After rotation, the code prints the updated array to the console.