Decimal to binary using c language - c

Question

I am very new to Coding. Here is a program that I wrote to convert decimal to binary
but there is one problem I am getting the result but it's in reverse
Example: Binary of 122 is 1111010 and I'm getting output 0101111.
Can anyone please tell me is it possible to reverse the output in my code?
Or what changes can I make in the following to get the correct output?
#include<stdio.h>
int main()
{
int n, q, r;
printf("Enter the decimal Number : ");
scanf("%d", &n);
int num=n;
while(n!=0)
{
r=n%2;
q=n/2;
printf("%d", r);
n=q;
}
return 0;
}

Answer 1

Seems like you are new to coding. It doesn't matter here is the problem.
Converting decimal to binary is like this,
eg:
division by 2
quotient
reminder
bit
10/2
5
0
0
5/2
2
1
1
2/2
1
0
2
1/2
0
1
3
=(1010)
So the output should have digits from bottom to top of the reminder column. Your output is printed from top to bottom.
See the code below where you need an array in order to store reminders and print the array in reverse order so you get the output you need
#include<stdio.h>
#include<stdlib.h>
int main(void){
int a[10],n,i;
printf("Enter the decimal Number : ");
scanf("%d",&n);
for(i=0;n>0;i++)
{
a[i]=n%2;
n=n/2;
}
printf("\nBinary of Given Number is=");
for(i=i-1;i>=0;i--)
{
printf("%d",a[i]);
}
return 0;
}

Answer 2

void display(unsigned n)
{
if(n == 0) return;
display(n /2);
printf("%d", n % 2);
}
and example usage:
https://godbolt.org/z/ahGPc74nf
As a homework: how to correctly handle 0?
Or not recursive version. This one can print or not leading zeroes:
void display(unsigned n, int printzeroes)
{
unsigned mask = 1 << (CHAR_BIT * sizeof(mask) - 1);
int print = printzeroes;
while(mask)
{
if(n & mask)
{
print = 1;
}
if(print) printf("%d", !!(n & mask));
mask >>= 1;
}
}
And usage: https://godbolt.org/z/7Eq71TMWb

Answer 3

First of all, please note that all numbers in a C program are to be regarded as binary. It's a common misconception among beginners that different number formats somehow co-exist in the executable program. But everything there is raw binary.
Sure the programmer may write numbers in different formats 7, 07 or 0x7 in the source code, but they get translated to binary by the compiler. Therefore, converting between binary and "x" doesn't make sense, because everything is already binary. You may however, convert from binary to a decimal string or similar, for the purpose of displaying a number to the user etc.
With that misconception out of the way - yes, you can create a binary string with the method you picked, dividing by ten and checking the remainder. The problem with that approach is that you'll get the most significant digit first. This is why you get the number backwards. So in order to do that, you'd have to store down the string in a character array first, before displaying it.
A more convenient way would be to use the "bitwise" operators like & and shift to mask out bit by bit in the data. Basically this:
if(n & (1u << bit)) // 1u to avoid shifting signed type
printf("1");
else
printf("0");
Where bit is the bit position 7,6,5... down to 0. If we prefer an up-counting loop instead, we can tweak the code into:
for(size_t i=0; i<8; i++)
{
size_t mask = 1u << 8-i-1;
...
}
And finally we can make the output a bit more compact, which is just a stylistic concern:
for(size_t i=0; i<8; i++)
{
size_t mask = 1u << 8-i-1;
printf("%c", n & mask ? '1' : '0');
}
If you aren't dead certain about C operator precedence, then use parenthesis, which is perfectly fine too:
for(size_t i=0; i<8; i++)
{
size_t mask = 1u << (8-i-1);
printf("%c", (n & mask) ? '1' : '0');
}

Answer 4

Here's the homework demanded by 0___________:
void printbin(int n)
{
static int depth;
if (n) ++depth, printbin(n / 2), --depth;
else if (depth) return; // print leading 0 only for n = 0
printf("%d", n % 2);
}
A leading 0 is not printed unless on the topmost recursion level.

Answer 5

You could maybe create an array of size 32 and keep adding the digits to the array. Or create a int variable and add the digits to the variable.
int result = 0;
while(n!=0)
{
r=n%2;
n=n/2;
result = result*10+r;
}