Array contains garbage values not input values - c

Question

I am writing a basic program to compute the binary eq of a decimal value. I'm storing the individual bits or 0 and 1 values into an array so I can eventually reverse the array and print the accurate binary representation. However when I print the array contents to check if array has been properly filled I see garbage values, or 0 if arr[]={0}
My code
int main() {
int i = 0, j = 0, k, decimal, binary = 0, remainder, divider;
int bin[10];
printf("Enter decimal value");
scanf("%d", &decimal);
while ((decimal != 0) && (i < decimal)) {
remainder = decimal % 2;
decimal = decimal / 2;
bin[i] = remainder;
j++;
printf("%d", bin[i]);
}
printf("\n%d", j);
printf("\n%d", bin[0]);
printf("\n%d", bin[1]);
printf("\n%d", bin[2]);
printf("\n%d", bin[3]);
printf("%d", bin);
return 0;
}
.exe
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Answer 1

If you are still having problems with the conversion, it may be helpful to consider a couple of points. First, you are over-thinking the conversion from decimal to binary. For any given integer value, the value is already stored in memory in binary.
For example, when you have the integer 10, the computer stores it as 1010 in memory. So for all practical purposes, all you need to do is read the memory for value and set your array values to 1 for each bit that is 1 and 0 for each bit that is 0. You can even go one better, since what you are most likely after is the binary representation of the number, there is no need to store the 1s and 0s as a full 4-byte integer value in bin, why not make bin a character array and store the characters '1' or '0' in the character array (which when nul-terminated) allows a simple printing of the binary representation as a string.
This provides several benefits. Rather than converting from base 10 to base 2 and the divisions and modulo calls required for the base conversion, you can simply shift decimal to the right by one and check whether the least-significant-bit is 0 or 1 and store the desired character '0' or '1' based on the results of a simple unary and operation.
For example, in you case with an integer, you can determine the number of bits required to represent any integer value in binary with sizeof (int) * CHAR_BIT (where CHAR_BIT is a constant provided in limits.h and specifies the number of bits in a character (e.g. byte)). For an integer you could use:
#include <stdio.h>
#include <limits.h> /* for CHAR_BIT */
#define NBITS sizeof(int) * CHAR_BIT /* constant for bits in int */
To store the character representations of the binary number (or you could store the integers 1, 0 if desired), you can simply declare a character array:
char bin[NBITS + 1] = ""; /* declare storage for NBITS + 1 char */
char *p = bin + NBITS; /* initialize to the nul-terminating char */
(initialized to all zero and the +1 to allow for the nul-terminating character to allow the array to be treated as a string when filled)
Next, as you have discovered, whether you perform the base conversion or shift and and the resulting order of the individual bit values will be in reverse order. To handle that, you can simply declare a pointer pointing to the last character in your array and fill the array with 1s and 0s from the back toward the front.
Here too the character array/string representation makes things easier. Having initialized your array to all zero, you can start writing to your array beginning at the next to last character and working from the end to the beginning will insure you have a nul-terminated string when done. Further, regardless of the number of bits that make up decimal, you are always left with a pointer to the start of the binary representation.
Depending on how you loop over each bit in decimal, you may need to handle the case where decimal = 0; separately. (since you loop while there are bits in decimal, the loop won't execute if decimal = 0;) A simple if can handle the case and your else can simply loop over all bits in decimal:
if (decimal == 0) /* handle decimal == 0 separately */
*--p = '0';
else /* loop shifting decimal right by one until 0 */
for (; decimal && p > bin; decimal >>= 1)
*--p = (decimal & 1) ? '1' : '0'; /* decrement p and set
* char to '1' or '0' */
(note: since p was pointing to the nul-terminating character, you must decrement p with the pre-decrement operator (e.g. --p) before dereferencing and assigning the character or value)
All that remains is outputting your binary representation, and if done as above, it is a simple printf ("%s\n", p);. Putting all the pieces together, you could do something like the following:
#include <stdio.h>
#include <limits.h> /* for CHAR_BIT */
#define NBITS sizeof(int) * CHAR_BIT /* constant for bits in int */
int main (void) {
int decimal = 0;
char bin[NBITS + 1] = ""; /* declare storage for NBITS + 1 char */
char *p = bin + NBITS; /* initialize to the nul-terminating char */
printf ("enter a integer value: "); /* prompt for input */
if (scanf ("%d", &decimal) != 1) { /* validate ALL user input */
fputs ("error: invalid input.\n", stderr);
return 1;
}
if (decimal == 0) /* handle decimal == 0 separately */
*--p = '0';
else /* loop shifting decimal right by one until 0 */
for (; decimal && p > bin; decimal >>= 1)
*--p = (decimal & 1) ? '1' : '0'; /* decrement p and set
* char to '1' or '0' */
printf ("binary: %s\n", p); /* output the binary string */
return 0;
}
(note: the comment on validating ALL user input -- especially when using the scanf family of functions. Otherwise you can easily stray off into Undefined Behavior on an accidental entry of something that doesn't begin with a digit)
Example Use/Output
$ ./bin/int2bin
enter a integer value: 0
binary: 0
$ ./bin/int2bin
enter a integer value: 2
binary: 10
$ ./bin/int2bin
enter a integer value: 15
binary: 1111
Two's-complement of negative values:
$ ./bin/int2bin
enter a integer value: -15
binary: 11111111111111111111111111110001
Look things over and let me know if you have any questions, or if you really need bin to be an array of int. Having an integer array holding the individual bit values doesn't make a whole lot of sense, but if that is what you have to do, I'm happy to help.