How to print binary array as characters

Question

I have an array that holds 28 ints which are all 1's and 0's. However, I need to print this information as 4 characters so how do I get each 7 bytes of data to become one bit in order to print.

Not sure this makes sense so I will illustrate what I need to:

Right now my array (in order) is this: 0101101111011101011000100010 But I need to somehow take those first 7 numbers (0101101) and print that out as Z and do that with the next 7, the next 7...

Thanks for your help!

Take your input 0101101. If you left pad that by 0, you get 00101101 = 0x2D = '-'. If you right pad that by 0, you get 01011010 = 0x5A = 'Z'. That is counter intuitive. Is that what you really want, to pad a zero to the right? — R Sahu
– R Sahu, Commented Apr 14, 2014 at 4:32
@R Sahu Bits are to be interpreted in little endian, not big. — chux
– chux, Commented Apr 14, 2014 at 4:44
The first 2 groups of 7 looks like "Zw", but that last 2 groups are not "+Z". — chux
– chux, Commented Apr 14, 2014 at 4:51
@chux Is that a general practice? I would love to read up on that topic, if you have a reference handy. — R Sahu
– R Sahu, Commented Apr 14, 2014 at 5:09

taylorcressy · Accepted Answer · 2014-04-14 13:20:39Z

1

I think this might be something along the lines you are looking for.

int to_int(int *bits) {
   int power = 2;
   int digit = 1;
   int value = 0;

   int i=0;
   for(i=0; i <= 6; i++) {
        if(bits[i] == 1) {
            value += digit;
        }
        digit *= power;
    }

    return value;
}  


int main() {
    int myArray[28] = {0, 1, 0, 1, 1, 0, 1,
                    1, 1, 1, 0, 1, 1, 1,
                    0, 1, 0, 1, 1, 0, 0,
                    0, 1, 0, 0 ,0, 1, 0};

    char theChars[5];
    theChars[0] = to_char(&myArray[0]);
    theChars[1] = to_char(&myArray[7]);
    theChars[2] = to_char(&myArray[14]);
    theChars[3] = to_char(&myArray[21]);
    theChars[4] = '\0';
    printf("%s\n",&theChars[0]);
}

Also, I don't think your expected output is correct.

edited Apr 14, 2014 at 13:20

answered Apr 14, 2014 at 2:28

taylorcressy

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Comments

Jamato · Accepted Answer · 2014-04-14 00:51:36Z

0

Well, there is always the stupid way: Cycle through each 7 blocks.

int bytes=7;
for(int i=0; i++;i<4){
     double ASCII = 0;
     for(int j=0; i++;j<bytes){
     ASCII+=Math.pow(2, bytes-j-1)*array[i*bytes + j]
     }
     char c = (char) ASCII // you'll have some trouble with types here
}

answered Apr 14, 2014 at 0:51

Jamato

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Comments

Chris Ryding · Accepted Answer · 2014-04-14 01:04:08Z

0

Assuming your input array is called inputBits[] Try something like this:

const int input_bit_count = 28;
char output[input_bit_count / 7];
int outIdx = 0;

// step through the bit stream converting bits to 7-bit characters
for( int inIdx = 0; inIdx < input_bit_count; ){
    // shift over and add the next bit to this character
    output[outIdx] <<= 1;
    if( inputBits[inIdx] != 0 ){
        output[outIdx] |= 1;
    }

    inIdx++;
    if( inIdx % 7 == 0)
        // after each 7 bits, increment to next output character
        outIdx++;
}

// done processing, now print it out
for( int chIdx = 0; chIdx < input_bit_count / 7; chIdx++ ){
    printf( "%c", output[chIdx] );
}

answered Apr 14, 2014 at 1:04

Chris Ryding

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3 Comments

user3366369 Over a year ago

When I try that my output is: AMv3 My output is supposed to be: Zw+Z"

Chris Ryding Over a year ago

So then what have you tried? Perhaps if you post some code we can help you debug it. Also, I simply had the bit order backwards, so change the if statement to if( inputBits[input_bit_count - inIdx - 1] != 0 ).

chux Over a year ago

1) output[] is not initialized to 0, thus output[outIdx] <<= 1; is shifting random data. 2) I suspect the bits are little endian. This solution assume big.

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