How to convert to binary as string in code C

Question

I am trying to convert int to binary as string but I can not. Please help me. How to convert integer to binary, please tell me.

Input: 32
Output: 00100000

My code:

#include <stdio.h>
#include <string.h>
char converttobinary(int n)
{
    int i;
    int a[8];
    char op;
    for (i = 0; i < 8; i++)
    {
        a[i] = n % 2;
        n = (n - a[i]) / 2;
    }
    for (i = 7; i >= 0; i--)
    {
        op = strcat(op, a[i]);
    }
    return op;
}
int main()
{
    int n;
    char str;
    n = 254;
    str = converttobinary(n);
    printf("%c", str);
    return 0;

}

So. what type is op and which values can it have? What is the return type of your function and which values can it return? — Eugene Sh.
– Eugene Sh., Commented Jul 10, 2019 at 14:50
You need to go back to your textbook or tutorial and study the difference between characters and strings. — Barmar
– Barmar, Commented Jul 10, 2019 at 15:24

Jonathan Leffler · Accepted Answer · 2019-07-10 16:13:28Z

1

I have tried to modify your solution with minimal changes to make it work. There are elegant solutions to convert Integer to Binary for example using shift operators.

One of the main issue in the code was you were using character instead of character array. i.e char str; instead of char str[SIZE]; Also you were performing string operations on a single character. Additionally, iostream header file is for C++.

There is room for lot of improvements in the solution posted below (I only made your code work with minimal changes).

My suggestion is to make your C basics strong and approach this problem again.

#include <stdio.h>
#include <string.h>

void converttobinary(int n, char *op)
{
    int i;
    int a[8];
    for (i = 0; i < 8; i++)
    {
        a[i] = n % 2;
        n = (n - a[i]) / 2;
    }
    for (i = 7; i >= 0; i--)
    {
        op[i]=a[i];
    }

}
int main()
{
    int n,i;
    char str[8];
    n = 8;
    converttobinary(n,str);
    for (i = 7; i >= 0; i--)
    {
        printf(" %d ",str[i]);
    }
    return 0;

}

edited Jul 10, 2019 at 16:13

Jonathan Leffler

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answered Jul 10, 2019 at 15:40

j23

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1 Comment

Jonathan Leffler Over a year ago

When is the string null-terminated? Well, it isn't null-terminated, so it isn't actually a string — by definition, a string is null-terminated. You could use printf("%8.8s\n", str); to safely print the byte array without spaces between the digits.

0___________ · Accepted Answer · 2019-07-10 15:47:45Z

0

char *rev(char *str)
{
    char *end = str + strlen(str) - 1;
    char *saved = str;
    while(end > str)
    {
        int tmp = *str;
        *str++ = *end;
        *end-- = tmp;
    }
    return saved;
}

char *tobin(char *buff, unsigned long long data)
{
    char *saved = buff;

    while(data)
    {
        *buff++ = (data & 1) + '0';
        data >>= 1;
    }
    *buff = 0;
    return rev(saved);
}

int main()
{

    char x[128];
    unsigned long long z = 0x103;
    printf("%llu is 0b%s\n", z, tobin(x, z));

    return 0;
}

answered Jul 10, 2019 at 15:47

0___________

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Comments

Marc Aynes · Accepted Answer · 2019-07-10 16:29:15Z

0

I modify your code a little bit to make what you want, the result of this code with

n = 10

is

00001010

In this code i shift the bits n positions of the imput and compare if there is 1 or 0 in this position and write a '1' if there is a 1 or a '0' if we have a 0.

#include <stdio.h>
void converttobinary(int n, char op[8]){
    int auxiliar = n;
    int i;

    for (i = 0; i < 8; i++) {
        auxiliar = auxiliar >> i;
        if (auxiliar & 1 == 1){
            op[7-i] = '1';
        } else{
            op[7-i] = '0';
        }
        auxiliar = n;
    }

}

int main (void){
    int n = 10;
    int i;
    char op[8];

    converttobinary(n, op);

    for(i = 7; i > -1; i--){
        printf("%c",op[i]);
    }

    return 0;

}

edited Jul 10, 2019 at 16:29

answered Jul 10, 2019 at 16:12

Marc Aynes

333 bronze badges

2 Comments

Jonathan Leffler Over a year ago

The output 'string' isn't null terminated, so it isn't a string, and you can't safely print it with %s. You could use printf("%8.8s\n", op); to print the byte-array.

Marc Aynes Over a year ago

@JonathanLeffler thats true i just fix it thanks :)

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