The current version of office is "Office 19" most people use "Office 365" though.
A "Pdf" or "Portable Document Format."
Green computing is the environmentally responsible and eco-friendly use of computers and their resources. In broader terms, it is also defined as the study of designing, manufacturing/engineering, using and disposing of computing devices in a way that reduces their environmental impact.
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:
bears = {"Grizzly":"angry", "Brown":"friendly", "Polar":"friendly"}
for bear in bears:
if bears[bear] == "friendly":
print("Hello, "+bear+" bear!")
else:
print("odd")
Explanation:
A dictionary called bears is given. A dictionary consists of key-value pairs.
You need to check each key-value pairs in bears and find the ones that have "friendly" as value using a for loop and if-else structure. In order to access the values of the dictionary, use the dictionary name and the key (inside the loop, the key is represented as bear variable). If a key has a value of "friendly", print the greeting. Otherwise, print "odd".
