How to print binary from an integer - c

Question

#include <stdio.h>
#include <stdlib.h>
void main()
{
int testNums[] = {3, 0x12, 0xFF, -3};
int testBits[] = {9, 7, 12, 15};
int i;
for (i = 0; i < sizeof(testNums) / sizeof(testNums[0]); i++)
printBin(testNums[i], testBits[i]);
}
void printBin(int num, int bits)
{
int pow;
int mask = 1 << bits - 1;
for(pow=0; pow<bits; pow++)
{
if(mask & num)
printf("1");
else
printf("0");
num<<1;
}
printf("\n");
}
It doesn't print out the correct binary number, but has the right amount of bits, and advice on how I can fix this?

Answer 1

Your problem is here, in printBin:
num<<1;
This statement does nothing. What you want is:
num<<=1;
For clarity, you should also use parenthesis here:
int mask = 1 << (bits - 1);
And you should move printBin above main so the definition is visible at the time the function is called.
Finally, main should always return int.

Answer 2

There are several issues in your code:
You need a prototype for printBin().
Use int main(void) or int main(int argc, char** argv) instead of void main().
num<<1; is an expression that has no side effects. Perhaps you are able infer this from compiler's warning. Write num<<=1; to make a difference.
Only a suggestion: Why not replace the if statement with printf("%d", !!(mask & num));?
Here is the fixed code that compiles without warnings using clang:
#include <stdio.h>
#include <stdlib.h>
void printBin(int num, int bits);
int main(void)
{
int testNums[] = {3, 0x12, 0xFF, -3};
int testBits[] = {9, 7, 12, 15};
int i;
for (i = 0; i < sizeof testNums / sizeof testNums[0]; i++)
printBin(testNums[i], testBits[i]);
return 0;
}
void printBin(int num, int bits)
{
int pow;
int mask = 1 << (bits - 1);
for(pow=0; pow<bits; pow++)
{
printf("%d", mask & num);
num <<= 1;
}
putchar('\n');
}
Output:
000000011
0010010
000011111111
111111111111101

Answer 3

Although your approach is conventional and #dbush's answer is appropriate, I would like like to mention another approach. You may know this and it doesn't handle the negative values very well but in this approach, you can convert number of any base into other very easily
#include <stdio.h>
#include <string.h>
#define RESOLUTION 128
void compute(int number, int base, char* placeHolder){
int k = 0;
int i, j;
char temp;
while(number){
placeHolder[k++] = "0123456789ABCDEF"[number % base];
number /= base;
}
if (k != 0) {
placeHolder[k] = '\0';
for(i = 0, j = strlen(placeHolder) - 1; i < j; i++, j--){
temp = placeHolder[i];
placeHolder[i] = placeHolder[j];
placeHolder[j] = temp;
}
}else{
placeHolder[0] = '0';
placeHolder[1] = '\0';
}
}
int main(void) {
char bin[RESOLUTION];
compute(0xFF, 10, bin);
printf("%s\n", bin);
return 0;
}
Change the RESOLUTION value at you will.
More specific to binary, you can print binary of any datatype like this
void printBits(size_t const size, void const * const ptr)
{
unsigned char *b = (unsigned char*) ptr;
unsigned char byte;
int i, j;
for (i=size-1;i>=0;i--)
{
for (j=7;j>=0;j--)
{
byte = b[i] & (1<<j);
byte >>= j;
printf("%u", byte);
}
}
puts("");
}
int main()
{
int i = 23;
uint ui = UINT_MAX;
float f = 23.45f;
printBits(sizeof(i), &i);
printBits(sizeof(ui), &ui);
printBits(sizeof(f), &f);
return 0;
}