Understanding a program in C - c

Question

# include <stdio.h>
int main()
{
int n, c, k;
printf("Enter an integer in decimal number system\n");
scanf("%d", &n);
printf("%d in binary number system is:\n", n);
for (c = 31; c >= 0; c--)
{
k = n >> c;
if (k & 1)
printf("1");
else
printf ("0");
}
printf("\n");
return 0;
}
Can someone help me understand how does this code work? I found this in an online site. It gives the binary counterpart of the number when the decimal is entered. But I don't understand how it is working. And also what does the symbol n>>c mean? Sorry if this is too silly, I am just a beginner in C. Please help me if you can. Thanks a lot.

Answer 1

Okay:
We declare some integers for use later in the program.
int n, c, k;
We get an input number from the user. printf() will print the string to the screen, and scanf() will request data from the user and store it in n.
printf("Enter an integer in decimal number system\n");
scanf("%d", &n);
We now print back to the user the value of n
printf("%d in binary number system is:\n", n);
This is a for loop. It has three parts. The first part gets run before we enter the loop, but only on the first iteration of the loop. So we set c to 31. Then, we check the condition. Is c >= 0? Yes. It is.
We can ignore the last part for now.
for (c = 31; c >= 0; c--)
{
This assigns to k the value of n shifted to the right c times. This is similar to dividing a number by 2 and then removing any decimals. So 5 >> 2 == 1. And 7 >> 1 == 3. Many calculator programs have this button as well.
This iteration, we have n as whatever the user input (Let's say 52). and c is 31. So this would have a value of 0 this loop.
k = n >> c;
Now, this test functions as an even or odd test. If it is odd, we print 1, and 0 otherwise. This also can be seen as looking at the bottom bit of a number and choosing if it is 1 or 0.
if (k & 1)
printf("1");
else
printf ("0");
We now finish this block of code, and we go back to the for loop. Now, on following iterations, we will execute the third block of code (the --c). So c == 30. Then, we will run the middle section, and verify that we still meet the conditions to execute this loop.
}
I imagine the remainder of the program is not interesting.
printf("\n");
return 0;
}
Now, what does this code actually do?
So this code as a whole attempts to print the binary representation of a number. It assumes a few things, including that the number fits in an integer, and the system's integers are 32 bits.
More generally, in each loop of the for loop, we grab the last bit of the number that we haven't seen yet, and print 1 if that bit is high.
If you don't know binary notation yet, then that is something that you really need to learn to have this code make any sense.

Answer 2

The statement
k = n >> c;
right shift n to c bits. The expression k & 1 in if returns 1 if the last bit of k is 1 else it returns 0. Taking n = 8 and c = 7, see the example (assuming 8 bit representation):
iteration print
--------- -----
1. k = 0000 1000 >> 7 = 0000 0000
0000 0000 & 0000 0001 = 0000 0000 = 0 ==> 0
2. k = 0000 1000 >> 6 = 0000 0000
0000 0000 & 0000 0001 = 0000 0000 = 0 ==> 0
3. k = 0000 1000 >> 5 = 0000 0000
0000 0000 & 0000 0001 = 0000 0000 = 0 ==> 0
4. k = 0000 1000 >> 4 = 0000 0000
0000 0000 & 0000 0001 = 0000 0000 = 0 ==> 0
5. k = 0000 1000 >> 3 = 0000 0001
0000 0001 & 0000 0001 = 0000 0000 = 0 ==> 1
. ..
. ..
8. k = 0000 1000 >> 0 = 0000 0000
0000 0000 & 0000 0001 = 0000 0000 = 0 ==> 0