Question

 #include <stdio.h> 
#include <stdlib.h>

// Print out the binary value of an integer

void binval (int num);
int get_number();

main () {
  int num;

  num = get_number();
  printf("Num = %d\n",num);
  binval(num);

}

void binval (int num) {
  int val = 0;
  int test;

  if (!num) {
    printf("\n");
    return;
  }


  test = num & 0x0001;
  if (test == 1) {
    val = 1;
 }

  num = num / 2;
  printf("%d",val);
  binval(num);
  return;
}

int get_number() { 
  int value = 0;
  char c;
  printf("Please input number : ");
  while ((c = getchar()) != '\n') { 
    if ( (c>'9') || (c<'0') ) { 
      printf("Incorrect character entered as a number - %c\n",c);
      exit(-1);
    }
    else {
      value = 10*value + c - '0';
    }
  }
  return(value);
}

now it compiles just get 01 for every answer.

Answer 1

The call to binval in main implicitly declares it like int binval(int).

In order to fix that, you’ll need to either add a forward declaration for binval, or move main to after binval.

After that, and after fixing the other stuff mentioned in other answers, we still have one little problem: the bits will be printed out backwards, because you’re printing the least significant bit first. The easiest way to fix that is to switch the binval and printf within the binval function itself, but of course you probably won’t want the base case (num==0) to print a newline. Just print a newline after you call the function from main.