Answer:
ansList =input().split() #get input and split it by space
ansList = [int(i) for i in ansList if int(i)>0] #turn string to integer,and get all positive integers
print(ansList)
Explanation: I think this would work for you. I leace comments in the answer.
Answer:
The answer is open source
Explanation:
hope this helps
# look it up on google
Answer:
#include <iostream>
using namespace std;
void divide(int numerator, int denominator, int *quotient, int *remainder)
{
*quotient = (int)(numerator / denominator);
*remainder = numerator % denominator;
}
int main()
{
int num = 42, den = 5, quotient=0, remainder=0;
divide(num, den, "ient, &remainder);
return 0;
}
Explanation:
The exercise is for "Call by pointers". This technique is particularly useful when a variable needs to be changed by a function. In our case, the quotient and the remainder. The '&' is passing by address. Since the function is calling a pointer. We need to pass an address. This way, the function will alter the value at the address.
To sum up, in case we hadn't used pointers here, the quotient and remainder that we set to '0' would have remained zero because the function would've made copies of them, altered the copies and then DELETED the copies. When we pass by pointer, the computer goes inside the memory and changes it at the address. No new copies are made. And the value of the variable is updated.
Thanks! :)
Answer:
The printItem() method code is filled in the explanation, highlighted with bold font.
Explanation:
// ===== Code from file BaseItem.java =====
public class BaseItem {
protected String lastName;
public void setLastName(String providedName) {
lastName = providedName;
return;
}
// FIXME: Define printItem() method
/* Your solution goes here */
public void printItem() {
// TODO Auto-generated method stub
System.out.println("Last name: "+lastName);
}
}
// ===== end =====
