Given:char newFunc(int & a, int b){char output = 'Q';%3Dwhile (a

Program: //include the header file #include <iostream> using namespace std; //definition of…

Answered: Given: char newFunc(int & a, int b) {…

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

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#include using namespace std; const int y = 1; int main () { int static y = 2; int i = 3, j = 4, m = 5, n = 6; %3D int a = [] (int x, int i = 1) { return x * i; } (y, 3) ; %3D int b = [=] (int x) { return [=] (int b) { return b + j; } (x) % 7; } (a); int c = [=] (int x) mutable ->int { m = 6; return [6] (int j) mutable { y = a * b; return y / j; } (x) -m; } (b) ; cout << a << endl; cout << b << endl; cout << c << endl; cout << m << endl; cout << y < endl; return 0; This program outputs 5 lines. What are they? Please explain your work and your answer.
int func(int a, int b) { return (a
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;…
Please explain this question void main() {int a =300; char *ptr = (char*) &a ; ptr ++; *ptr =2; printf("%d", a); }
int sum = 0; for (int i 0; i < 5; i++){ sum += i; } cout << sum;
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; }
CFG: Example 1 • Draw the CFG for the following code: int f(int n){ } int m = n* n; if (n < 0) else return 0; return m;
int j; for (int i = 0; i < 5; i++) {j = 2*j*i;} What is j?
int a[10] = {0,1,2,3,4,5,6,7,8,9};int *m = &a[0];int *p = &a[5];int *q = &a[1]; p = (int) m + (int) p - (int) q; what is the value of of *p
1. #include int main() { int i; for (i = 1; i int main() { double number, sum = 0; do { printf("Enter a number: "); scanf("%lf", &number); sum += number; } while(number != 0.0); printf("Sum = %.2lf",sum); return 0; } }
Int cube ( int x){ // returns cube of x : return x.x.x. :)* O False True
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 {…

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Given: char newFunc(int & a, int b) { char output = 'Q'; %3D while (a