**Given the following code, what is the final value of variable `i`?**```cint i;for ( i = 0 ; i <= 4 ; i++ ){ printf("i is:\n", i);}```*Answer in the box below:* [__________]This code snippet is a simple `for` loop in C programming language. Here's a breakdown of what each part of the code does:- `int i;`: This line declares an integer variable `i`.- `for ( i = 0 ; i <= 4 ; i++ )`: This line initializes `i` to 0, sets the loop to continue executing as long as `i` is less than or equal to 4, and increments `i` by 1 after each iteration.- `{ printf("i is:\n", i); }`: This block of code will be executed in each iteration. It prints the current value of `i` to the console.- The loop starts with `i = 0` and continues until `i` exceeds 4. Therefore, the loop runs for `i` values 0, 1, 2, 3, and 4. After `i` is incremented from 4 to 5, the condition `i <= 4` becomes false, and the loop terminates.Thus, the final value of `i` after the loop completes is 5.

Given code, initially i declare. When for loop start i=0 For loop executive i=0 to i<= 4. I…

Answered: Given the following code, what is the…

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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io/courses/1443/quizzes/4601/take/questions/74256 Study the definition of the following two functions "Rate" and "Bonus" and answer the question below. int Rate (int one, int two) { int R = one; for (int i = 0; i 20) num2 num1 * 20 +R; %3D else numl = num2 * 10 + R; return num1 + num2; What is the output of the following code segment? cout << Bonus (3, 5); a hp 近

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#include using namespace std; main () { for (n = 5;n > e; n--) cout << n<<" "; int int n; %3D { if (n == 3) break; 543 53 554 O 567
1) Find the errors in the following codes: int main() { } int { Hanging Indent #include ; int a = 1;b = 4; c, d C + = a++; d = 5--; printf("Please enter an inter: "); scanf("%f", num) d = (num = 0? a+b:a+b+c); sum= (a:b:c:d: a+b+c+d); printf("Sum is %d: \n", Sum); Return 0;
Trace the following code and give the output int a = 2; cout << a; while (a < 200) { a *= 2; cout << ", " << a; }
#include <stdio.h>#include <stdlib.h>int LineFunc(void);int main(){int a, b;printf("Enter a :");scanf("%d", &a);printf("Enter b :");scanf("%d", &b);LineFunc();printf("\n\n");system("pause"); }int LineFunc(void){int i,j;int a=1, b = 0, y; for (i = -10; i < 11; i++){for (j = 10; j > -11; j--){y = a * i + b;if (i == 0 ){printf("-");}else if (j == 0){printf("|");}else if (i == 0 && j == 0){printf("+");}else if (y == j){printf("*");}elseprintf(" ");}printf("\n");} } Could you please write the code that gives different results for each value in the chart?
1) Find the errors in the following codes: int main() { int { Hanging Indent #include ; int a = 1;b = 4; c, d C + = a++; d = 5--; printf("Please enter an inter: "); scanf("%f", num) d = (num = 0? a+b:a+b+c); sum = (a:b:c:d: a+b+c+d); printf("Sum is %d: \n", Sum); Return 0; } 2) Assume a, b, c and n are float variables, and d, e, fand m are integer variables, what is the result of each “printf” function in the following program? Please do not run IDE to get the results.
#include<stdio.h> #include<math.h> int main() { char cod; int gal; float cost; printf("Data Codes:"); printf(" \n H-Home \n C-Commercial \n I-Industrial "); printf("\nInput data code: "); scanf("%c", &cod); printf("\nInput number of water (in gallons) used: "); scanf("%d", &gal); if (cod=='H'|| cod=='h') { cost = 250+(0.002*gal); printf("\nTOTAL COST IS %.2f", cost); } else if (cod=='C'||cod=='c') { if (gal<=4000000) { printf("\nTOTAL COST IS 5000.00"); }…
in the following code there is * #include #include int man() { int B=25; int *p; //b memory location=1234; p=&b; cout>> b; gtch(); return 0; 2 error no error 3 error 4 error
1-)Can you improve this code for a 2-digit number? For instance, 27 means twenty-seven? #include <stdio.h>#include <stdlib.h> int main() {int a;printf("Enter a integer number :");scanf("%d",&a); switch(a){case 0:printf("ZERO");break;case 1:printf("ONE");break;case 2:switchprintf("TWO");break;case 3:printf("THREE");break;case 4:printf("FOUR");break;case 5:printf("FIVE");break;case 6:printf("SIX");break;case 7:printf("SEVEN");break;case 8:printf("EIGHT");break;case 9:printf("NINE");break; } }
The rule for implicit type conversion in 'C' is A. int < unsigned < float < double B. unsigned < int < float < double C. int < unsigned < double < float D. unsigned < int < double < float
Refer to the statement below, #include void main() { int i; int number[11]={12,15,17,3,2,7,10,10,15,15,50}; for(i=0;i < 11; į++){ printf(" %d", number[i]); } Write a segment in C language to: Compute the average number Find the maхітит аnd minimum питber

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Given the following code, what is the final value of variable 1 ? int i; for ( i=® ; i <= 4 ; i++ ) {) printf("i is:\n", i);