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Answer:
The correct answer is:
C. ndx = 0;
while (ndx < 3) {
ar[ndx] = 0;
ndx++;
}
Explanation:
The declaration given is:
int ar[3];
This means the array consists of three locations and is named as ar.
We know that the indexes are used to address the locations of an array and the index starts from 0 and goes upto to 1 less than the size of the array which means the indexes of array of 3 elements will start from 0 and end at 2.
Now in the given options we are using ndx variable to run the while loop.
So the code to assign zero to all elements of array will be
ndx = 0;
while(ndx<3)
{
ar[ndx] = 0;
ndx++;
}
Hence, the correct answer is:
C. ndx = 0;
while (ndx < 3) {
ar[ndx] = 0;
ndx++;
}
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.
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The application program that saves data automatically as it is entered is the MS Access.
