Using atoi() in C to parse character array with binary values

Question

I'm trying to convert a string of binary characters to an integer value.

For example: "100101101001" I would split it into four segments using a for loop then store it in array[4]. However whenever I use the function atoi(), I encounter a problem where it does not convert the character string properly if the string starts with "0".

An example would be "1001" = 1001, but if it is 0110 it would be converted to 110, also with 0001 it would be come only 1.

Here is the code that I made:

for(i = 0; i < strlen(store); i++)
{
    bits[counter] = store [i];
    counter++;
    if(counter == 4)
    {   
        sscanf(bits, "%d", &testing);
        printf("%d\n", testing);
        counter = 0;
    }       
}

Answer 1

The atoi() function only converts decimal numbers, in base 10.

You can use strtoul() to convert binary numbers, by specifying a base argument of 2. There is no need to "split" the string, and leading zeroes won't matter of course (as they shouldn't, 00010₂ is equal to 10₂):

const char *binary = "00010";
unsigned long value;
char *endp = NULL;

value = strtoul(binary, &endp, 2);
if(endp != NULL && *endp == '\0')
  printf("converted binary '%s' to integer %lu\n", binary, value);

Answer 2

atoi() convert from char array to int and not to binary

you can use the following function

int chartobin(char *s, unsigned int *x) {
    int len = strlen(s), bit;
    *x = 0;
    if(len>32 || len<1) return -1;
    while(*s) {
        bit = (*s++ - '0');
        if((bit&(~1U))!=0) return -1;
        if (bit) *x += (1<<(len-1));
        len--;
    }
    return 0;
}

Tested and it works