Ok if lets say a young child at the age of like 6-7 was allowed to play grand theft auto by his/her parents maybe the child would understand incorrectly and start doing what the people do in the game this may cause to serious trouble jail time or even death. In this case dont let your kids play underrated games or watch underrated movies.
Hope that helped you understand more.
Answer:
error: incompatible types
Explanation:
Given
The attached code
Required
The output
Variable "a" is declared as float
While p is declared as a pointer to an integer variable
An error of incompatible types will be returned on line 3, <em>int *p = a;</em>
Because the variables are not the same.
To assign a to p*, we have to use type casting.
Hence, (b) is correct
Answer:
// code in C++
#include <bits/stdc++.h>
using namespace std;
// main function
int main()
{
// variables
int sum_even=0,sum_odd=0,eve_count=0,odd_count=0;
int largest=INT_MIN;
int smallest=INT_MAX;
int n;
cout<<"Enter 10 Integers:";
// read 10 Integers
for(int a=0;a<10;a++)
{
cin>>n;
// find largest
if(n>largest)
largest=n;
// find smallest
if(n<smallest)
smallest=n;
// if input is even
if(n%2==0)
{
// sum of even
sum_even+=n;
// even count
eve_count++;
}
else
{
// sum of odd
sum_odd+=n;
// odd count
odd_count++;
}
}
// print sum of even
cout<<"Sum of all even numbers is: "<<sum_even<<endl;
// print sum of odd
cout<<"Sum of all odd numbers is: "<<sum_odd<<endl;
// print largest
cout<<"largest Integer is: "<<largest<<endl;
// print smallest
cout<<"smallest Integer is: "<<smallest<<endl;
// print even count
cout<<"count of even number is: "<<eve_count<<endl;
// print odd cout
cout<<"count of odd number is: "<<odd_count<<endl;
return 0;
}
Explanation:
Read an integer from user.If the input is greater that largest then update the largest.If the input is smaller than smallest then update the smallest.Then check if input is even then add it to sum_even and increment the eve_count.If the input is odd then add it to sum_odd and increment the odd_count.Repeat this for 10 inputs. Then print sum of all even inputs, sum of all odd inputs, largest among all, smallest among all, count of even inputs and count of odd inputs.
Output:
Enter 10 Integers:1 3 4 2 10 11 12 44 5 20
Sum of all even numbers is: 92
Sum of all odd numbers is: 20
largest Integer is: 44
smallest Integer is: 1
count of even number is: 6
count of odd number is: 4
Answer:
The program is as follows:
#include <iostream>
#include <iomanip>
using namespace std;
int main(){
int num1, num2, num3;
cin>>num1>>num2>>num3;
cout << fixed << setprecision(2);
cout<<(num1 + num2 + num3)/3<<" ";
cout<<num1 * num2 * num3<<" ";
return 0;
}
Explanation:
This declares three integer variables
int num1, num2, num3;
This gets input for the three integers
cin>>num1>>num2>>num3;
This is used to set the precision to 2
cout << fixed << setprecision(2);
This prints the average
cout<<(num1 + num2 + num3)/3<<" ";
This prints the product
cout<<num1 * num2 * num3<<" ";
