2 iii) int a = 5; //L1 //L2 for(int i = 5; i <= 6; i++) { а %3D а + 3; System.out.println(a); //L3 //L4 a = a + 2; //L5
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Explain in your own words what the code in 2iii does step by step.
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Solved in 2 steps with 1 images
- int sum = 0; for (int i 0; i < 5; i++){ sum += i; } cout << sum;For the C++ program show on the next page, answer the questions that follow. The names g, j, x, and y are declared multiple times. Distinguish uses of these names using line numbers where they were declared as subscripts. a. At point A in the program, what names are in scope? b. At point A in the program, what variables are live? c. At point B in the program, what names are in scope? d. At point B in the program, what variables are live?debug
- int main(){ long long int total; long long int init; scanf("%lld %lld", &total, &init); getchar(); long long int max = init; long long int min = init; int i; for (i = 0; i < total; i++) { char op1 = '0'; char op2 = '0'; long long int num1 = 0; long long int num2 = 0; scanf("%c %lld %c %lld", &op1, &num1, &op2, &num2); getchar(); long long int maxr = max; long long int minr = min; if (op1 == '+') { long long int sum = max + num1; maxr = sum; minr = sum; long long int res = min + num1; if (res > maxr) { max = res; } if (res < minr) { minr = res; } } else { long long int sum = max * num1; maxr = sum; minr = sum; long long int res = min * num1;…Number of digits int main() {int c=0;int i=3856; int l=i; while(l>0) {l=l/10; c++; } printf("%d",c); return 0; } Summation of digits int main() {int sum=0; int i=384; int l=i; while(l>0) { sum+=l%10; l=l/10;} printf("%d",sum); return 0; }Q1 function myFunc() { let a = 10; if (true) { Q3 } } let a = 5; console.log(a); Q4 console.log(a); Q2 const square = num => num * num; console.log(square(3) + 5); myFunc(); let x = 30; let y "200"; console.log(x+y); const nums = [10, 20, 8, 17]; nums.filter(e=> e > 10); console.log(nums); Q5 const nums = [30, 35, 42, 20, 15]; console.log(nums.every (num => num > 20)); January 15نقطة واحدة Let A = {a; b; c; d} and R= {(a; a); (b; c); (c; b); (d; d)} then R is Transitive Equivalent not transitiveFind errors / syntax error. Write line number#include using namespace std; class fraction{ float d,s; public: fraction() { d=0; s=1; fraction(int a, int b) { d=%3; s=b3; } void print() { cout<int pertambahan(int pilihan, int now){ int result; result =pilihan + now; return result;} int perkalian(int pilihan, int now){ int result; result = pilihan*now; return result;} int main(){ int NoR,initial; int now; int temp1, temp2; scanf("%d %d", &NoR, &initial); getchar(); now=initial; for(int o=0;o<NoR;o++){ char a,b; int pilihan1,pilihan2; scanf("%c %d %c %d",&a, &pilihan1,&b, &pilihan2); getchar(); if(a=='+') { if(pilihan1>=-1000 && pilihan1<=1000) { temp1 = pertambahan(pilihan1,now); } else{ temp1=-10000; } } else if(a=='x') { if(pilihan1>=-2 && pilihan1<=2){ temp1 = perkalian(pilihan1,now); } else { temp1=-10000; } } if(b=='+') { if(pilihan2>=-1000 && pilihan2<=1000) { temp2 = pertambahan(pilihan2,now); } else {…SEE MORE QUESTIONSRecommended textbooks for youC++ Programming: From Problem Analysis to Program…Computer ScienceISBN:9781337102087Author:D. S. MalikPublisher:Cengage LearningC++ Programming: From Problem Analysis to Program…Computer ScienceISBN:9781337102087Author:D. S. MalikPublisher:Cengage Learning