why is the output of this code 00 10 #include main( ){ int i,j; for(i=0; i<2; i++){ for(j=0; j<2; j++){ printf("%d %d\n",i,j); break; } } }
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why is the output of this code 00 10
#include<stdio.h>
main( ){
int i,j;
for(i=0; i<2; i++){
for(j=0; j<2; j++){
printf("%d %d\n",i,j);
break;
}
}
}
Step by step
Solved in 2 steps with 1 images
- Void Do1 (int: &, a. int &b) { a = 5; a = a + b; b = a + 2; } Int main() { Int x = 10; Do1 (x,x); Cout << x << endl; } The output of this program is#include int main() { } int a, b, c, i=0; for (c=0; c < 7; C++) { } for(b 0; b < c; b++) { } for(a 0; a < b; a++) { } return 0; shown above? printf("%d\n", i++); ) What is the final value of i in the code#include <stdio.h> int main(){ int M; printf("Enter the number of chairs you want to enter:\n"); scanf("%d", &M); double price[M]; double sum = 0.0; int i=0; while(i < M) { printf("Enter price of chair #%d: ",(i+1)); scanf("%lf", &price[i]); if(price[i] < 1003) { price[i] = price[i] - price[i] * 0.10; sum = sum + price[i]; } else if(price[i] >= 1003 && price[i] <= 2006) price[i] = price[i] - price[i] * 0.15; else if(price[i] > 2006) price[i] = price[i] - price[i] * 0.20; printf("Price of chair #%d after discount is %.3lf\n", (i+1), price[i]); i++; } printf("\n\nTotal price after discount of all chairs which got a discount of 10%% is "); printf("%.3lf", sum); int N = 2; printf("\n\nN is equal to the sum of My EUI ID(1+1) = 2"); double p[N]; printf("\nEnter price of %d desks: ",…
- #include<iostream>using namespace std;void main(){double pi = 0, denominator = 1;int counter = 999999;for (int x = 0; x < counter; x++){if (x % 2 != 0){pi = pi - (1 / denominator);}else{pi = pi + (1 / denominator);}denominator = denominator + 2;}pi = pi * 4;cout << " So the computed value of a PI is = " << pi << endl;cout << " ";//return 0;system("pause");} Note: This a program called ComputePI to compute the value of π Tutor just have to Modify This program to use nested-if (if ... else if ... else if ... else) instead. Explain by applying a double line comment#include <stdio.h>void cubeByReference( int *nPtr ); // function prototypeint main( void ){ int number = 5; // initialize number printf("The original value of number is %d", number ); // pass address of number to cubeByReference cubeByReference( &number ); printf("\nThe new value of number is %d\n", number );} // end main void cubeByReference( int *nPtr ){ *nPtr = *nPtr* *nPtr* *nPtr;} passing argument by reference - We modify the code above 1- define a second argument (example "int number2 = 9") and a pointer to it 2- define a second function (addByReference) that adds number2 to number - passing both arguments by reference 3- print-out the result (that is in number) Upload the output and .c codeint func(int a, int b) { return (aCFG: 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;#include <stdio.h>int main(){int d;int s[20],i, r, p, lg=0,m;char c;printf("Enter number of salesman(max 20): ");scanf("%d", &d);for(i=0; i<d; i++){ printf("\n salesman %d sales: ");scanf("%d", &s[i]);}for(i=0; i<d; i++){for(r=i+1; j<d; r++){if(s[i] > s[r]){p= s[i];s[i] = s[r];s[j] = p;}}}printf("\nsalesman lowest to highest: ");for(i=0; i<d; i++){printf("%d\t", s[i]);}for(i=0;i<m;i++){printf("\n highest sales: %d ",lg);if(lg<=s[i])lg=s[i];break;}getch();} >>> the upper part output should be like this enter number of salesman (max 20): 5 salesman 1 500 salesman 2 300 salesman 3 1000 salesman 4 200 salesman 5 1000 in the lower part the lowest to highest the output should become like this salesman 4 200 salesman 2 300 salesman 1 500 salesman 3 1000 salesman 5 1000 highest total sales : 2000io/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#include using namespace std; int main () int x, y; { X = 5; y = ++x cout << x << y; %3D ++x; %3D x = 5; y = x++ * ++x; cout << x < y; return e; } 525636 747953 749735 636636Please explain this question void main() {int a =300; char *ptr = (char*) &a ; ptr ++; *ptr =2; printf("%d", a); }SEE MORE QUESTIONSRecommended textbooks for youDatabase System ConceptsComputer ScienceISBN:9780078022159Author:Abraham Silberschatz Professor, Henry F. Korth, S. SudarshanPublisher:McGraw-Hill Education
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