Thread: Can someone explain this program to me?

Code:

#include <stdio.h>

int left(int a);
int right(int b);

int main(void)
{

int n = 1, x = 1;

	while( n < 50 )
		{
			printf("%3d, %3d\n", n, x);
				x = left( n );
				n = right( x );
		}
	printf("%3d, %3d\n", n, x);
}
	int left(int a)
		{
			return (a * 2);
		}
	int right(int b)
		{
			return (b * 3);
		}

Code:

 Output:
1,   1
6,   2
36,  12
216,  72
Press any key to continue . . .

I'm trying to understand how this function program works. I dont understand how 'n' increases to be more than 50, but from I can tell by messing around with the program, is that when the printf in the 'while' is taken out, the program only prints the last set of numbers '216, 72'. I see the pattern with the numbers, but I have no clue how the program does it.

Both x and n increase when they are assigned the values returned by the functions left() and right(). A greater value is produced in that function and is then assigned to either x or n in main.

For example:

x = left(n) = n*2 = 2.

Then n = right(x) = x * 3 = 2 * 3 = 6.

Ok, I think I understand it now. So there arent any real values for a & b, and instead left and right represent 2 & 3.
So basically in order, the program goes,

The program prints "1 , 1" for the values of " n, x" which are defined outside the 'while'
It then calculates x = (2)(1), where left (a) returns as 2
n = (3)(2), where right (b) returns as 3
The program prints "6 , 2" for the values of " n, x" which are now defined within the 'while'
It then calculates x = (2)(6)
n = (3)(12), where 12 comes from the 2 * 6
The program prints "36 , 12"
It then calculates x = (2)(36)
n = (3)(72)
The program prints " 216 , 72"

Is that a correct interpretation?

Yes, remember argument names are just generic names for the function. It is their call that determines their actual value inside the function. So by calling left(x) you are basically indicating to the function to use the value of x for a.