C++ Programming - Functions - Discussion
Discussion Forum : Functions - Programs (Q.No. 33)
33.
What will be the output of the following program?
#include<iostream.h>
struct IndiaBix
{
int arr[5];
public:
void BixFunction(void);
void Display(void);
};
void IndiaBix::Display(void)
{
for(int i = 0; i < 5; i++)
cout<< arr[i] << " " ;
}
void IndiaBix::BixFunction(void)
{
static int i = 0, j = 4;
int tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp ;
i++;
j--;
if(j != i) BixFunction();
}
int main()
{
IndiaBix objBix = {{ 5, 6, 3, 9, 0 }};
objBix.BixFunction();
objBix.Display();
return 0;
}Discussion:
2 comments Page 1 of 1.
Divya said:
8 years ago
Please explain this.
Anthony Dawson said:
8 years ago
The key here is i & j are local static to IndiaBix::BixFunction().
On calling IndiaBix::BixFunction() the first time i = 0 and j = 4. This causes the first call of IndiaBix::BixFunction() to swap the 1st and last elements of the array. i.e. {0, 6, 3, 9, 5}. Then increment i and decrement j such that i = 1 and j = 3. The statement j != i is therefore true and IndiaBix::BixFunction() is called again.
On the second call to IndiaBix::BixFunction() i = 1 and j = 3 because they are static. This causes the second call of IndiaBix::BixFunction() to swap the 2nd and 4th elements of the array. i.e. {0, 9, 3, 6, 5}. Then increment i and decrement j such that i = 2 and j = 2. This time the statement j != i is false so that is the end of the recursion.
This leaves IndiaBix::Display() to print "0 9 3 6 5 ".
Hope this helps.
On calling IndiaBix::BixFunction() the first time i = 0 and j = 4. This causes the first call of IndiaBix::BixFunction() to swap the 1st and last elements of the array. i.e. {0, 6, 3, 9, 5}. Then increment i and decrement j such that i = 1 and j = 3. The statement j != i is therefore true and IndiaBix::BixFunction() is called again.
On the second call to IndiaBix::BixFunction() i = 1 and j = 3 because they are static. This causes the second call of IndiaBix::BixFunction() to swap the 2nd and 4th elements of the array. i.e. {0, 9, 3, 6, 5}. Then increment i and decrement j such that i = 2 and j = 2. This time the statement j != i is false so that is the end of the recursion.
This leaves IndiaBix::Display() to print "0 9 3 6 5 ".
Hope this helps.
(1)
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