Trouble printing decimal to binary number properly - c

Question

I'm trying to write code to print the binary representation of a number. This was my attempt:
#include <stdio.h>
void getBinary(int num);
int main(void) {
int num = 2;
getBinary(num);
return 0;
}
void getBinary(int num) {
int mask = 1 << 31, i;
int temp;
for (i = 31; i >= 0; i--) {
temp = num & mask;
if (temp > 0)
printf("1");
else
printf("0");
mask = mask >> 1;
}
printf("\n");
}
And this doesn't work. If i make num = 2 my output will be ...0011. This is the answer I was given:
void getBinary(int);
int main()
{
int num=0;
printf("Enter an integer number :");
scanf("%d",&num);
printf("\nBinary value of %d is =",num);
getBinary(num);
return 0;
}
/*Function definition : getBinary()*/
void getBinary(int n)
{
int loop;
/*loop=15 , for 16 bits value, 15th bit to 0th bit*/
for(loop=15; loop>=0; loop--)
{
if( (1 << loop) & n)
printf("1");
else
printf("0");
}
}
and this will give the correct output. Why? What is different between my code and this. Aren't they doing the same thing with the solution doing it in fewer steps?

Answer 1

#include <stdio.h>
void getBinary(int num);
int main(void) {
unsigned int num = 2;
getBinary(num);
return 0;
}
void getBinary(int num) {
unsigned int mask = 1 << 31;
int i;
unsigned int temp;
for (i = 31; i >= 0; i--) {
temp = num & mask;
if (temp > 0)
printf("1");
else
printf("0");
mask = mask >> 1;
}
printf("\n");
}
For those curious, this is the correct answer. Just make the mask unsigned.

Answer 2

You have some problems in the code:
1 << 31 invokes undefined behavior because of signed arithmetic overflow. On your platform it most likely produces the value INT_MIN, that has the expected bit configuration but is negative.
temp > 0 might fail incorrectly if num < 0, because the sign bit will make temp negative too.
shifting mask to the right when its value is negative has an implementation defined result: On your platform it duplicates the sign bit, therefore mask will not have a single bit as expected, but all bits set above the current one. This explains the observed behavior.
You should use type unsigned int in function getBinary for num, temp and mask to get correct behavior.
Here is a corrected version of your code:
#include <stdio.h>
void getBinary(unsigned int num);
int main(void) {
unsigned int num = 2;
getBinary(num);
return 0;
}
void getBinary(unsigned int num) {
unsigned int mask = 1U << 31;
unsigned int temp;
for (int i = 31; i >= 0; i--) {
temp = num & mask;
if (temp != 0)
printf("1");
else
printf("0");
mask = mask >> 1;
}
printf("\n");
}
The proposed solution only prints 16 bits, which may or may not have been the specification. If the int type is larger than 16 bits, the bit shift will not overflow and everything is fine. If int has 16 bits, 1 << 15 invokes undefined behavior, so the solution is not strictly correct, but will function on most current platforms.
Here is a simpler solution:
void getBinary(unsigned int num) {
for (int shift = 32; shift-- > 0;) {
putchar('0' + ((num >> shift) & 1));
}
putchar("\n");
}
Initialize shift to 16 to print only the low order 16 bits.