12.14 Zylab 3 - Single Procedure Call Given an array of at least one integer, write a program to create a new array with elements equal to the power of each element in the original array raised to the index, i.e., P[i] = A[i]^i. For this, write two functions that will be called in main function independently. • power o inputs: element (A[i]) and index (1) • task: returns the value of element raised to index (A[i]^i). } • newElement o inputs: base address of new array P (*P), current size of P (variable k) and the new element (A[i]^i) • task: add the new element at the end. o This function does not return any value (void). Following is a sample C code to perform the required task. You may modify the code for the functions, but the task performed should not be changed. int main() { // Variable int *A, *P; int n, k; int pow; Declaration // Base addresses of A and P // Lengths of arrays A and B // Return value from power function // Task of main function P[0] = 1; } k++; // 0th element = A[0]^0 = 1 for (int j = 1; j < n; j++) { k = j; // Current length of array B pow power (A[j], j); newElement(P, k, pow); int power (int a, int b) { int pow= a;
12.14 Zylab 3 - Single Procedure Call Given an array of at least one integer, write a program to create a new array with elements equal to the power of each element in the original array raised to the index, i.e., P[i] = A[i]^i. For this, write two functions that will be called in main function independently. • power o inputs: element (A[i]) and index (1) • task: returns the value of element raised to index (A[i]^i). } • newElement o inputs: base address of new array P (*P), current size of P (variable k) and the new element (A[i]^i) • task: add the new element at the end. o This function does not return any value (void). Following is a sample C code to perform the required task. You may modify the code for the functions, but the task performed should not be changed. int main() { // Variable int *A, *P; int n, k; int pow; Declaration // Base addresses of A and P // Lengths of arrays A and B // Return value from power function // Task of main function P[0] = 1; } k++; // 0th element = A[0]^0 = 1 for (int j = 1; j < n; j++) { k = j; // Current length of array B pow power (A[j], j); newElement(P, k, pow); int power (int a, int b) { int pow= a;
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
Related questions
Question
![12.14 Zylab 3 - Single Procedure Call
Given an array of at least one integer, write a program to create a new array with elements equal to the power of each element in the
original array raised to the index, i.e., P[i] = A[i]^i.
For this, write two functions that will be called in main function independently.
● power
inputs: element (A[i]) and index (1)
• task: returns the value of element raised to index (A[i]^i).
}
O
• newElement
inputs: base address of new array P (*P), current size of P (variable k) and the new element (A[i]^i)
o task: add the new element at the end.
o This function does not return any value (void).
O
Following is a sample C code to perform the required task. You may modify the code for the functions, but the task performed should not be
changed.
int main() {
// Variable Declaration
int *A, *P;
int n, k;
int pow;
// Task of main function
P[0] 1;
for (int j =
k = j;
pow
// Base addresses of A and P
// Lengths of arrays A and B
// Return value from power function
}
k++;
// 0th element = A[0]^0 = 1
1; j < n; j++) {
// Current length of array B
power (A[j], j);
newElement(P, k, pow);
int power(int a, int b) {
int pow= a;](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F909b55a2-117c-4c26-8933-e5a86ba10ee3%2F01e10ccc-8b4e-4c44-aa35-c0c254aa1e29%2F1xz9h8_processed.png&w=3840&q=75)
Transcribed Image Text:12.14 Zylab 3 - Single Procedure Call
Given an array of at least one integer, write a program to create a new array with elements equal to the power of each element in the
original array raised to the index, i.e., P[i] = A[i]^i.
For this, write two functions that will be called in main function independently.
● power
inputs: element (A[i]) and index (1)
• task: returns the value of element raised to index (A[i]^i).
}
O
• newElement
inputs: base address of new array P (*P), current size of P (variable k) and the new element (A[i]^i)
o task: add the new element at the end.
o This function does not return any value (void).
O
Following is a sample C code to perform the required task. You may modify the code for the functions, but the task performed should not be
changed.
int main() {
// Variable Declaration
int *A, *P;
int n, k;
int pow;
// Task of main function
P[0] 1;
for (int j =
k = j;
pow
// Base addresses of A and P
// Lengths of arrays A and B
// Return value from power function
}
k++;
// 0th element = A[0]^0 = 1
1; j < n; j++) {
// Current length of array B
power (A[j], j);
newElement(P, k, pow);
int power(int a, int b) {
int pow= a;
![int main() {
}
}
// Variable Declaration
int *A, *P;
int n, k;
int pow;
}
// Task of main function
P[0]
1;
=
}
k++;
// Base addresses of A and P
// Lengths of arrays A and B
// Return value from power function
// 0th element
for (int j = 1; j < n; j++) {
k = j;
}
return (pow);
=
pow power (A[j], j);
newElement(P, k, pow);
int power(int a, int b) {
int pow= a;
for (int 1 1; 1 < b; 1++) {
pow = pow * a;
// Current length of array B
A[0]^0 = 1
void newElement(int* P, int k, int pow) {
P[k] = pow;
Registers Variables
$50
A
$s1
$s2
$s3
n
P
k](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F909b55a2-117c-4c26-8933-e5a86ba10ee3%2F01e10ccc-8b4e-4c44-aa35-c0c254aa1e29%2Fvjmxq0y_processed.png&w=3840&q=75)
Transcribed Image Text:int main() {
}
}
// Variable Declaration
int *A, *P;
int n, k;
int pow;
}
// Task of main function
P[0]
1;
=
}
k++;
// Base addresses of A and P
// Lengths of arrays A and B
// Return value from power function
// 0th element
for (int j = 1; j < n; j++) {
k = j;
}
return (pow);
=
pow power (A[j], j);
newElement(P, k, pow);
int power(int a, int b) {
int pow= a;
for (int 1 1; 1 < b; 1++) {
pow = pow * a;
// Current length of array B
A[0]^0 = 1
void newElement(int* P, int k, int pow) {
P[k] = pow;
Registers Variables
$50
A
$s1
$s2
$s3
n
P
k
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