converting integer into 4 bit binary

2
votes

I have the following code to convert an integer into a binary representation for each integer from 0 to 15, thus I need 4-bit representation of an integer.

the code it works fine but the problem is the length of the binary is not correct, so when given 1 as int it returns 1 as output instead of 0001. for 2 it should return 0010 but instead, it returns 10.

How can I change this code so it returns the correct representation? Using printf %04d is oki when printing the results, but just for printing because the actual value is still different.

I was trying to make another method that gets an integer convert it into a string and then depending on its length add 0 before it until the length is 4.

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

int convert(int dec)
{
    if (dec == 0)
    {
        //printf("Base c\n");
        return 0;
    }
    else
    {
        //printf("Rec call\n");
        return (dec % 2) + 10 * convert(dec / 2);
    }
}

int main(int argc, char *argv[]) 
{
    // argc is number of arguments given including a.out in command line
    // argv is a list of string containing command line arguments

    int v = atoi(argv[2]);
    printf("the number is:%d \n", v);


    int bin = 0;

    if(v >= 0 && v <= 15){
        printf("Correct input \n");
        bin = convert(v);
        printf("Binary is: %04d \n", bin);
        printf("Binary is: %d \n", bin);

    }
    else{
        printf("Inorrect input, number cant be accepted");
    }

What I need this method to do: 1. given an integer 2. return 4-bit representation of this integer, not sure if it should be an int or string. 3. for example convert(2) should return 0010, 6 returns 110, and I want this to be 0110 and so on. the method convert was supposed to do that for me. I hope I am clear about what I what to happen.

this should return for:

1 0001
2 0010
3 0011
4 0100
5 0101
6 0110
and so on
15 1111
3
cbinary

3 Answers

2
votes

Your requirements are very unclear but from the comments I think I have gathered what you require. According to my suggestion, you need to change the function to return a string instead of an int.

You need to pass an argument in which the string should be returned. So the function will be -

char * convert(int dec, char *output) {
    output[4] = '\0';
    output[3] = (dec & 1) + '0';
    output[2] = ((dec >> 1) & 1) + '0';
    output[1] = ((dec >> 2) & 1) + '0';
    output[0] = ((dec >> 3) & 1) + '0';
    return output;
}

This function can be used as 

 char binary[5];
 convert(15, binary);
 printf("%s", binary);

Demo: Ideone

1
votes

Initialize a character array of size 5, ex: char arr[5] and mark arr[4] ='\0', rest of the slots as 0 and then store the LSB into the 4th array slot & MSB near the 1st array slot. Print it using %s format specifier.

Initially the array will look like |0|0|0|0|\0|. Let's say that your input is 5 and the output you receive is (in form of int) 101, then after inserting the output into the array will look like |0|1|0|1|\0|.

0
votes

My attempt

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



int convert(int dec)
{
    if (dec == 0)
    {
        //printf("Base c\n");
        return 0;
    }
    else
    {
        //printf("Rec call\n");
        return (dec % 2) + 10 * convert(dec / 2);
    }
}
int main()
{
 int binary = 10;
 binary = convert(binary);
 printf("%d", binary);
 return 0;
}