How to print the binary representation of a character?

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How to print the binary representation of a character?

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#1 by (3 votes)

2

I need to get the binary representation of the characters of a string .

I can do the hexadecimal representation with the printf itself using %x .

Something like:

void imprime_hex(char *input) {
  for(int i = 0; i < strlen(input); i++) {
    printf("%x\n", input[i]);
  }
}

See running on Repl.it

Is there something similar to print the binary representation? Or some function that can help me get that result?

Btw, I imagine this will vary depending on encoding, etc. There's no need to worry about it, the idea is to use only the "common" characters: A-Z, a-z, 0-9.

c string

asked by anonymous 26.03.2018 / 23:05

1 answer

More than one list in the same loop Error calling function on my page

score 3 · Accepted Answer

Nothing is ready.

In fact to do something production ready needs to deal with a lot, analyze endianess , etc. Something naive not universal would be this:

#include <stdio.h>
#include <stdlib.h>

void BinFormat(char character, char *text) {
    text[8] = 'int divisor = (int)pow(2, i); //base 2 elevado à posição que está para achar o divisor
printf("[%d, ", divisor);
int cabe = character / divisor; //acha quantas unidades cabem no divisor
printf("%d, ", cabe);
int impar = (cabe % 2 != 0); //o que cabe é impar?
printf("%d] ", impar);
text[i] = impar + 48; //avança na tabela ASCII 48 posições para chegar em '0' ou '1'
';
    for (int i = 7; i >= 0; i--) text[i] = ((character >> i) & 1) + '0';
}

int main(void) {
    char *text = malloc(9);
    BinFormat('A', text);
    printf("%s\n", text);
    BinFormat('B', text);
    printf("%s\n", text);
    BinFormat('C', text);
    printf("%s\n", text);
}

See running on ideone . And no Coding Ground . Also put it in GitHub for future reference .

The strategy of how to allocate memory can vary. I do not really like to allocate in heap , but it's the simplest and most correct.

Because C has the incredible shortcoming of not being able to return an array , it could return a struct with 8 or 9 chars , but would then require a cast to transform into a real string .

Depending on the case you could have all the printable characters (95) in a static table, it would occupy 855 bytes, but would have the advantage of being very fast. But something more universal would require more memory.

No good solution.

So is it a more readable solution?

int divisor = i; //base 2 elevado à posição que está para achar o divisor
printf("[%d, ", divisor);
int cabe = character >> i; //acha quantas unidades cabem no divisor
printf("%d, ", cabe);
int impar = cabe & 1; //o que cabe é impar?
printf("%d] ", impar);
text[i] = impar + '0'; //avança na tabela ASCII até o '0' ou '1' se for 0 ou 1

See running ideone . Also put it in GitHub for future reference .

Let's now leave more C:

#include <stdio.h>
#include <stdlib.h>

void BinFormat(char character, char *text) {
    text[8] = 'int divisor = (int)pow(2, i); //base 2 elevado à posição que está para achar o divisor
printf("[%d, ", divisor);
int cabe = character / divisor; //acha quantas unidades cabem no divisor
printf("%d, ", cabe);
int impar = (cabe % 2 != 0); //o que cabe é impar?
printf("%d] ", impar);
text[i] = impar + 48; //avança na tabela ASCII 48 posições para chegar em '0' ou '1'
';
    for (int i = 7; i >= 0; i--) text[i] = ((character >> i) & 1) + '0';
}

int main(void) {
    char *text = malloc(9);
    BinFormat('A', text);
    printf("%s\n", text);
    BinFormat('B', text);
    printf("%s\n", text);
    BinFormat('C', text);
    printf("%s\n", text);
}