将十进制数字的每个数字转换为相应的二进制

Question

I need to convert the string "12345678" to the value 00010010001101000101011001111000 (the value in binary only without the zeroes on the left). So I have written this code in c, the problem is that when I run it does nothing, just waits like there is an error until I stop it manually. Any ideas?

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

void reduce(char string[]) {
    int i=0, j=0, k=0, cnt=0, tmp=4, num;
    char arr[4], result[4*strlen(string)];
    for (i=0; i<strlen(string); i++) {
            num = atoi(string[i]);
            while (num != 0) {
                    arr[j++] = num%2;
                    num = num/2;
                    tmp--;
            }
            while (tmp != 0) {
                    arr[j++] = 0;
                    tmp--;
            }
            j--;
            for (k=i*4; k<(i*4+4); k++) {
                    result[k++] = arr[j--];
            }
            j = 0;
            tmp = 4;
    }
    printf("The result is: \n");
    for (i=0; i<4*strlen(result); i++) {
            printf("%d",result[i]);
    }
    printf("\n");
}

 int main() {
    char c[8] = "12345678";
    reduce(c);
    return 0;
}

Answer 1

Lots of small errors in your code, which makes it hard to pin-point a single error. Main problem seems to be you are confusing binary numbers ( 0 , 1 ) with ASCII digits ( "0" , "1" ) and are mis-using string functions.

1. as mentioned elsewhere, char c[8] = .. is wrong.
2. atoi(string[i]) cannot work; it expects a string , not a char . Use `num = string[i]-'0';
3. arr[..] gets the value 'num%2 , that is, a numerical value. Better to use '0'+num%2 so it's a character string.
4. you increment k in result[k++] inside a loop that already increments k
5. add result[k] = 0; at the end before printing, so strlen works correctly
6. 4*strlen(result) is way too much -- the strlen is what it is.
7. you might as well do a simple printf("%s\\n", result);

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

void reduce(char string[]) {
    int i=0, j=0, k=0, cnt=0, tmp=4, num;
    char arr[5], result[4*strlen(string)+1];
    for (i=0; i<strlen(string); i++) {
            num = string[i]-'0';
            while (num != 0) {
                    arr[j++] = '0'+num%2;
                    num = num/2;
                    tmp--;
            }
            while (tmp != 0) {
                    arr[j++] = '0';
                    tmp--;
            }
            arr[j] = 0;
            j--;
            for (k=i*4; k<(i*4+4); k++) {
                    result[k] = arr[j--];
            }
            j = 0;
            tmp = 4;
    }
    result[k] = 0;
    printf("The result is: \n");
    for (i=0; i<strlen(result); i++) {
            printf("%c",result[i]);
    }
    printf("\n");
}

 int main() {
    char c[] = "12345678";
    reduce(c);
    return 0;
}

.. resulting in

The result is: 
00010010001101000101011001111000

Answer 2

In your main() , do either

char c[ ] = "12345678";

or

char c[9] = "12345678";

if you want to use c as a string. Otherwise, it does not have enough space to store the terminating null character.

Here, I took the liberty to modify the code accordingly to work for you. Check the below code. Hope it's self-explanatoty.

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

void reduce(char string[]) {
        int i=0, j=0, k=0, cnt=0,  count = 0;           //count added, tmp removed
        char arr[4], result[ (4*strlen(string) )+ 1], c;    //1 more byte space to hold null
        for (i=0; i<strlen(string); i++) {
                c = string[i];
                count = 4;
                while (count != 0) {                        //constant iteration 4 times baed on 9 = 1001
                        arr[j++] = '0' + (c%2);            //to store ASCII 0 or 1 [48/ 49]
                        c = c/2;
                        count--;
                }
/*      //not required
                while (tmp >= 0) {
                        arr[j++] = 0;
                        tmp--;
                }
*/
                j--;
                for (k=(i*4); k<((i*4) +4); k++) {
                        result[k] = arr[j--];
                }
                j = 0;
                memset (arr, 0, sizeof(arr));
        }
        result[k] = 0;
        printf("The result is: %s\n", result);   //why to loop when we've added the terminating null? print directly.
/*
        for (i=0; i< strlen(result); i++) {
                printf("%c",result[i]);
        }
        printf("\n");
*/
}

int main() {
        char c[ ] = "12345678";
        reduce(c);
        return 0;
}

Output:

[sourav@broadsword temp]$ ./a.out 
The result is: 00010010001101000101011001111000

Answer 3

It seems from your example that the conversion you are attempting is to binary coded decimal rather than binary . That being the case your solution is somewhat over-complicated; you simply need to convert each digit to its integer value then translate the bit pattern to ASCII 1 's and 0 's.

#include <stdio.h>

void reduce( const char* c )
{
    for( int d = 0; c[d] != 0; d++ )
    {
        int ci = c[d] - '0' ;

        for( unsigned mask = 0x8; mask != 0; mask >>= 1 )
        {
            putchar( (ci & mask) == 0 ? '0' : '1' )   ;
        }
    }
}

On the other hand if you did intend a conversion to binary (rather than BCD), then if the entire string is converted to an integer , you can directly translate the bit pattern to ASCII 1 's and 0 's as follows:

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


void reduce( const char* c )
{
    unsigned ci = (unsigned)atoi( c ) ;
    static const int BITS = sizeof(ci) * CHAR_BIT ;

    for( unsigned mask = 0x01 << (BITS - 1); mask != 0; mask >>= 1 )
    {
        putchar( (ci & mask) == 0 ? '0' : '1' )   ;
    }
}

Answer 4

1. Convert your string to an integer using int num = atoi(c) .

2. Then do

    int binary[50]; int q = num,i=0; while(q != 0) { binary[i++] = q%2; q = q/2; } 

Printing your binary array is reverse order will have your binary equivalent. Full program:

#include<stdio.h>


    int main(){

        char c[100];
        int num,q;

        int binary[100],i=0,j;


        scanf("%d",c);

        num = atoi(c);
        q = num;


        while(q!=0){

             binary[i++]= q % 2;

             q = q / 2;

        }




        for(j = i -1 ;j>= 0;j--)

             printf("%d",binary[j]);


        return 0;

    }

Answer 5

You can use the below reduce function.

void reduce(char string[])
{
    unsigned int in = atoi(string) ;
    int i = 0, result[32],k,j;

    while (in > 0) {
        j = in % 10; 
        k = 0;
        while (j > 0) {
            result[i++] = j % 2;
            j = j >> 1;
            k++;
        }
        while (k < 4) {
            result[i++] = 0;
            k++;
        }
        in = in/10; 
    }
    printf("Result\n");
    for(--i;i >= 0; i--) {
        printf("%d", result[i]);
    }
    printf("\n");

}

For 12345678 the output would be 00010010001101000101011001111000 , where each character is printed in its binary format.

Answer 6

It might need some adjustments, but it does the job as it is.

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

int
main(void)
{
        int  i;
        int  n;
        char *str       = "12345678";
        const int  bit  = 1 << (sizeof(n)*8 - 1);

        n = atoi(str);

        for(i=0; i < sizeof(n)*8 ; i++, n <<= 1)
                n&bit ? printf("1") : printf("0");

        return 0;
}