How to convert char value to binary and display each bit separately?

Question

As a practice exercise I'm trying to convert an entered char value to binary and display each bit of the 8 bit output. I have found some example code for conversion:

int main (void)
{
char a =16;
int i=0;
for (i = 0; i < 8;i++) { 
printf("%d", !!((a << i) & 0x80));}
return 0;
}

however when I use an entered value for a such as with the code:

int main(void){
char a;
printf("Enter char value:", a);
scanf("%c", &a);
printf("a=%c", a);
return 0 ;}

the incorrect value is given. And also how would I be able to display each bit separately? for example

printf("The sign bit is %d\n", &);
printf("The bit 6 is %d\n", &);
printf("The bit 5 is %d\n", &);
printf("The bit 4 is %d\n", &);
printf("The bit 3 is %d\n", &);
printf("The bit 2 is %d\n", &);
printf("The bit 1 is %d\n", &);
printf("The bit 0 is %d\n", &);

What do you want to happen with the `a` here? `printf("Enter char value:", a);` — Yunnosch, Apr 25 '18 at 05:28
If you init `a` with a recognisable value, is that what gets printed? — Yunnosch, Apr 25 '18 at 05:29
If hardcoding and entering as input yield different results, have you considered checking whether anything there behaves unexpectedly instead of blaming the conversion to binary? — Ulrich Eckhardt, Apr 25 '18 at 05:31
the entered value should be converted to binary and display each bit from sign bit to bit 0 — J.Smith, Apr 25 '18 at 05:32
I understood that, but what is supposed to happen to the variable which gets used in a printf which does not have any reference to any parameter in the format string? — Yunnosch, Apr 25 '18 at 05:33
https://stackoverflow.com/questions/7863499/conversion-of-char-to-binary-in-c follow this link — , Apr 25 '18 at 05:34
the value is already binary... ALL is binary in computer... You should use `unsigned char` for binary shift. — Stargateur, Apr 25 '18 at 05:34

score 0 · Answer 1 · answered Apr 25 '18 at 05:51

You could do a conversion doing math with the c code.. I did not click on the possible duplicate post but something like..

int i;
char c; 
int bits[8];
/*get a character from stdin and save it to c*/
i =  c + ‘0’;
if(i>=128){
    bits[7]=1;
     i= i-128;
}
if(i>=64){
    bits[6]=1;
    i = i - 64;
}
if(i >= 32){
    bits[5]=1;
    i =i- 32
}

// do this for all the bits

This should work and then you have bits[0] - bits[7] your 0-7 bits!! I am sure there is a better way to do this but this should work!!

`i = c + ‘0’;` why?? It's wrong. Further - for this to work use `unsigned char` and also `int bits[8] = {0};` to get the array initialized. Correcting that may make this work but why not use a simple loop? — Support Ukraine, Apr 25 '18 at 05:57

How to convert char value to binary and display each bit separately?

1 Answers1