May you help me trace this C program. Why is the first output 2?
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May you help me trace this C program. Why is the first output 2?
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- convert c code to mips these 2 functions are linked with each other. #define MAX_BOARD_SIZE 12 // Players #define PLAYER_EMPTY 0 #define PLAYER_BLACK 1 #define PLAYER_WHITE 2 int board_size; int current_player = PLAYER_BLACK; char board[MAX_BOARD_SIZE][MAX_BOARD_SIZE]; int main(void); void announce_winner(void); unsigned int count_discs(int player); void announce_winner(void) { int black_count = count_discs(PLAYER_BLACK); int white_count = count_discs(PLAYER_WHITE); if (white_count > black_count) { printf("The game is a win for WHITE!\n"); white_count += count_discs(PLAYER_EMPTY); } else if (black_count > white_count) { printf("The game is a win for BLACK!\n"); black_count += count_discs(PLAYER_EMPTY); } else { printf("The game is a tie! Wow!\n"); } printf("Score for black: %d, for white: %d.\n", black_count, white_count); } unsigned int count_discs(int player) { int count = 0; for (int row = 0; row < board_size; ++row) { for (int col = 0; col < board_size; ++col) { if…void fact(int num) { int k,f=1; for (k=1;k<=num; k++) f=f*k; cout<< f; int main() { int arri[3]={3,2,1} ; for(int i=0;i<3;i++) fact(arri[i]); } 261 O 621 O 7 462int machineCompare1(struct machine* mac1, struct machine* mac2){ struct mac1; struct mac2; char make1[21], make2[21]; char model1[51], model2[21]; if((lap1 -> make1 == lap2 -> make2) && (lap1 -> model1 == lap2 -> model2)) return 0; else return 1;} have to create function to compare make and model of 2 machines - it's using structures - not sure if this is correct/ on the right track?
- 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 }…#include<bits/stdc++.h>#include<math.h>using namespace std; class TotalResistance{double series_res,parallel_res,sp_res;public:TotalResistance(){series_res=parallel_res=sp_res=0;}void seriesResistance(double resistance[],int n);void parallelResistance(double resistance[],int n);void spResistance(double resistance[],int n);};void TotalResistance::seriesResistance(double resistance[],int n){for(int i=0;i<n;i++)series_res += resistance[i];cout<<"Total Resistance in series is: "<<series_res<<endl;}void TotalResistance::parallelResistance(double resistance[],int n){double temp=0;for(int i=0;i<n;i++)temp += (1/resistance[i]);parallel_res = 1/temp;cout<<"Total Resistance in parallel is: "<<parallel_res<<endl;}void TotalResistance::spResistance(double resistance[],int n){for(int i=0;i<n;i++)series_res += resistance[i];double temp=0;for(int i=0;i<n;i++)temp += (1/resistance[i]);parallel_res = 1/temp;cout<<"Total Resistance in…#include<bits/stdc++.h>#include<math.h>using namespace std; class TotalResistance{double series_res,parallel_res,sp_res;public:TotalResistance(){series_res=parallel_res=sp_res=0;}void seriesResistance(double resistance[],int n);void parallelResistance(double resistance[],int n);void spResistance(double resistance[],int n);};void TotalResistance::seriesResistance(double resistance[],int n){for(int i=0;i<n;i++)series_res += resistance[i];cout<<"Total Resistance in series is: "<<series_res<<endl;}void TotalResistance::parallelResistance(double resistance[],int n){double temp=0;for(int i=0;i<n;i++)temp += (1/resistance[i]);parallel_res = 1/temp;cout<<"Total Resistance in parallel is: "<<parallel_res<<endl;}void TotalResistance::spResistance(double resistance[],int n){for(int i=0;i<n;i++)series_res += resistance[i];double temp=0;for(int i=0;i<n;i++)temp += (1/resistance[i]);parallel_res = 1/temp;cout<<"Total Resistance in…
- int func(int a, int b) { return (a4. A={3,4}, B={4,5}. An B= AUB= A-B= A OB= AXB=Question 37 public static void main(String[] args) { Dog[] dogs = { new Dog(), new Dog()}; for(int i = 0; i >>"+decision()); } class Counter { private static int count; public static void inc() { count++;} public static int getCount() {return count;} } class Dog extends Counter{ public Dog(){} public void wo(){inc();} } class Cat extends Counter{ public Cat(){} public void me(){inc();} } The Correct answer: Nothing is output O 2 woofs and 5 mews O 2 woofs and 3 mews O 5 woofs and 5 mews O#include using namespace std; bool isPalindrome(int x) { int n=0,val; val = x; while(x > 0) { n = n * 10 + x % 10; x = x / 10; } } int main() { int n; cin >>n; if(isPalindrome(n)) { cout <JAVA LANGUAGE CODE Postfix Calculator by CodeChum Admin One good implementation of computing infix expressions is to transform them to postfix and then evaluate via the postfix expression. Infix expressions is the common way of writing arithmetic expressions. The binary operator come between them as shown below: 2 * 5 + 9 - 10 / 20 In postfix expressions, the operands come first before the operator: 2 5 * 9 + 10 20 / - A stack can be used to evaluate postfix expressions. The operands are pushed onto the Stack and when an operator is found two operands are popped and the operation is performed and finally the result is pushed back onto the Stack. The final answer will be the lone element of the Stack. Input The first line contains a positive integer n representing the number of postfix expressions. What follows are n postfix expressions themselves. 5 10 20 + 30 15 2 * + 100 20 30 + / 90 20 10 + + 0 / 9 3 - 10 + 2 * Output A single line containing the result of…int sum = 0; for (int i 0; i < 5; i++){ sum += i; } cout << sum;SEE MORE QUESTIONS