Print binary of an ascii character - c

Question

I am using the following approach to print the binary of an ascii character:
unsigned char bitmask = 0b10000000;
char _my_char = 'a';
for (int i=0; i<8; i++) {
if (i==4) putchar(' '); // separate every four characters
printf("%d", (_my_char & bitmask) != 0);
bitmask = bitmask >> 1;
}
putchar('\n');
return 0;
How could I generalize this so that, for example, I could print the binary of any data type/structure? One byte per row?

Answer 1

There are numerous ways to do this. The following outputs a nice chart of printable characters giving the value for each character in ASCII, decimal, hex and binary, e.g.
#include <stdio.h>
#include <limits.h>
/* CHAR_BIT */
#ifndef CHAR_BIT
#define CHAR_BIT 8
#endif
int main (void) {
char c = 0; /* character */
int i = 0; /* int loop counter */
printf ("\nchar | int | hex | binary\n");
printf ("-----+-------+--------+---------\n");
for (c = 32; c <= 126; c++) { /* for each char in printable range */
/* output the character, decimal, hex and binary */
printf (" %c | %3d | 0x%02x | ", c, c, c);
for (i = sizeof (c) * CHAR_BIT - 1; i >= 0; i--) /* for each bit */
printf ("%d", ((c >> i) & 0x1) ? 1 : 0); /* output 0/1 */
putchar ('\n'); /* output \n */
}
putchar ('\n');
}
Example Use/Output
$ ./bin/bin_ascii
char | int | hex | binary
-----+-------+--------+---------
| 32 | 0x20 | 00100000
! | 33 | 0x21 | 00100001
" | 34 | 0x22 | 00100010
# | 35 | 0x23 | 00100011
$ | 36 | 0x24 | 00100100
% | 37 | 0x25 | 00100101
& | 38 | 0x26 | 00100110
' | 39 | 0x27 | 00100111
( | 40 | 0x28 | 00101000
) | 41 | 0x29 | 00101001
* | 42 | 0x2a | 00101010
+ | 43 | 0x2b | 00101011
, | 44 | 0x2c | 00101100
- | 45 | 0x2d | 00101101
. | 46 | 0x2e | 00101110
/ | 47 | 0x2f | 00101111
0 | 48 | 0x30 | 00110000
1 | 49 | 0x31 | 00110001
2 | 50 | 0x32 | 00110010
3 | 51 | 0x33 | 00110011
4 | 52 | 0x34 | 00110100
5 | 53 | 0x35 | 00110101
6 | 54 | 0x36 | 00110110
7 | 55 | 0x37 | 00110111
8 | 56 | 0x38 | 00111000
9 | 57 | 0x39 | 00111001
<snip>
p | 112 | 0x70 | 01110000
q | 113 | 0x71 | 01110001
r | 114 | 0x72 | 01110010
s | 115 | 0x73 | 01110011
t | 116 | 0x74 | 01110100
u | 117 | 0x75 | 01110101
v | 118 | 0x76 | 01110110
w | 119 | 0x77 | 01110111
x | 120 | 0x78 | 01111000
y | 121 | 0x79 | 01111001
z | 122 | 0x7a | 01111010
{ | 123 | 0x7b | 01111011
| | 124 | 0x7c | 01111100
} | 125 | 0x7d | 01111101
~ | 126 | 0x7e | 01111110
Let me know if you have any questions about the logic.

Answer 2

You could create a function which takes a void pointer and the size of the structure as input.
void printdata(void *data, size_t len)
{
char my_char;
for (size_t m = 0; m < len; m++)
{
my_char = *(((char*) data) + m);
for (size_t i=0; i<8; i++)
{
if (i==4) printf(" "); // separate every four characters
printf("%d", (my_char & bitmask) != 0);
bitmask = bitmask >> 1;
}
printf("\n");
}
}
Then you can call this function for each element of your struct.
e.g.
typedef struct
{
int a;
char b;
int c[20];
}my_struct_t;
my_struct_t mystruct;
..
printdata(mystruct.a,sizeof(int));
printdata(mystruct.b,sizeof(char));
printdata(mystruct.c, 20*sizeof(int));
You can also pass the entire struct to the above function, but it will also print the padding bytes which is not what is needed.

Answer 3

This fit types what you want
#include <stdio.h>
#include <stdlib.h>
#define bitmask 0b1
void F1(void *a,int Len){
int i,j;
unsigned char*b=(unsigned char*)a;
b+=Len-1;
for(i=0;i<Len;i++,b--){
for(j=7;j>-1;j--){
printf("%d",((*b)&bitmask<<j)>>j);
}
}
printf("\n");
}
int main(){
unsigned int Q=10;
F1(&Q,sizeof(unsigned int));
unsigned short QQ=10;
F1(&QQ,sizeof(unsigned short));
unsigned char QQQ=10;
F1(&QQQ,sizeof(unsigned char));
return 0;
}