Print a Char Array of Integers in C - c

Question

For class, I am required to create a function that converts an Integer into it's corresponding Binary number. However, I am forced to use the given main and parameters for the to_binary function. The whole problem requires me to print out the 32 bit binary number, but to break it up, I am just trying to print out the Char Array, that I thought I filled with Integers (perhaps the issue). When I do compile, I receive just a blank line (from the \n) and I am wondering how I can fix this. All I want to do is to be able to print the binary number for 5 ("101") yet I can't seem to do it with my professor's restrictions. Remember: I cannot change the arguments in to_binary or the main, only the body of to_binary. Any help would be greatly appreciated.
#include<stdio.h>
void to_binary(int x, char c[]) {
int j = 0;
while (x != 0) {
c[j] x = x % 2;
j++;
}
c[33] = '\0';
}
int main() {
int i = 5;
char b[33];
to_binary(i,b);
printf("%s\n", b);
}

Answer 1

This is the answer to your question.
void to_binary(int x, char c[]) {
int i =0;
int j;
while(x) {
/* The operation results binary in reverse order.
* so right-shift the entire array and add new value in left side*/
for(j = i; j > 0; j--) {
c[j] = c[j-1];
}
c[0] = (x%2) + '0';
x = x/2;
i++;
}
c[i]=0;
}

Answer 2

the problem is in the code below:
while (x != 0) {
c[j] = x % 2; // origin: c[j] x = x % 2; a typo?
j++;
}
the result of x % 2 is a integer, but you assigned it to a character c[j] —— integer 1 is not equal to character '1'.
If you want to convert a integer(0-9) to a character form, for example: integer 7 to character '7', you can do this:
int integer = 7;
char ch = '0' + integer;

Answer 3

One of the previous answers has already discussed the issue with c[j] x = x % 2; and the lack of proper character conversion. That being said, I'll instead be pointing out a different issue. Note that this isn't a specific solution to your problem, rather, consider it to be a recommendation.
Hard-coding the placement of the null-terminator is not a good idea. In fact, it can result in some undesired behavior. Imagine I create an automatic char array of length 5. In memory, it might look something like this:
Values = _ _ _ _ _
Index = 0 1 2 3 4
If I were to populate the first three indexes with '1', '0', and '1', the array might look like so:
Values = 1 0 1 _ _
Index = 0 1 2 3 4
Let's say I set index 4 to contain the null-terminator. The array now looks like so:
Values = 1 0 1 _ \0
Index = 0 1 2 3 4
Notice how index three is an open slot? This is bad. In C/C++ automatic arrays contain garbage values by default. Furthermore, strings are usually printed by iterating from character to character until a null-terminator is encountered.
If the array were to look like it does in the previous example, printing it would yield a weird result. It would print 1, 0, 1, followed by an odd garbage value.
The solution is to set the null-terminator directly after the string ends. In this case, you want your array to look like this:
Values = 1 0 1 \0 _
Index = 0 1 2 3 4
The value of index 4 is irrelevant, as the print function will terminate upon reading index 3.
Here's a code example for reference:
#include <stdio.h>
int main() {
const size_t length = 4;
char binary[length];
size_t i = 0;
while (i < length - 1) {
char c = getchar();
if (c == '0' || c == '1')
binary[i++] = c;
}
binary[i] = '\0';
puts(binary);
return 0;
}

Answer 4

#include<stdio.h>
int binary(int x)
{
int y,i,b,a[100];
if(x<16)
{
if(x%2==1)
a[3]=1;
if(x/2==1||x/2==3 || x/2==5 || x/2==7)
a[2]=1;
if(x>4 && x<8)
a[1]=1;
else if(x>12 && x<16)
a[1]=1;
if(x>=8)
a[0]=1;
}
for(i=0;i<4;i++)
printf("\t%d",a[i]);
printf("\n");
}
int main()
{
int c;
printf("Enter the decimal number (less than 16 ):\n");
scanf("%d",&c);
binary(c);
}
this code might help it will simply convert the decimal number less than 16 into the 4 digit binary number.if it contains any error than let me know