Convert a String to Float

Code

#include <stdio.h>
#include <stdint.h>

float custom_atof(const char *str) {
    // Your logic here
    int i, center=0, left=0, right=0, sign=0;
    // look for a sign and for th edecimal point and for the end
    char *ptr;
    float result, multiplier;

    if (*str == '+') {
        sign = 1;
        left = 1;
    }
    else if (*str == '-') {
        sign = -1;
        left =1;
    }

    // look for the decimal point and the end
    i=0;
    center= -1;
    while (str[i] != '\0') {
        if (str[i] == '.') {
            center = i;
        }
        i++;
    }
    right = i;
    if (center == -1) {
        center = right;
    }

    //printf("sign: %d, left:%d, center:%d, right:%d\n", sign, left, center, right);

    result = 0.0;
    multiplier = 1.0;
    for (i=(center-1); i >= left; i-- ) {
        result += (str[i] - '0') * multiplier;
        multiplier *= 10.0;
    }

    multiplier = 0.1;
    for (i=(center+1); i < right; i++) {
        result += (str[i] -'0') * multiplier;
        multiplier *= 0.1;
    }

    if (sign == -1)
        result *= -1.0;


    return result;
}

int main() {
    char str[101];
    fgets(str, sizeof(str), stdin);

    // Remove newline
    uint8_t i = 0;
    while (str[i]) {
        if (str[i] == '\n') {
            str[i] = '\0';
            break;
        }
        i++;
    }

    float value = custom_atof(str);
    printf("%.2f", value);
    return 0;
}

Solving Approach

use counters to scan for sign decimal point and end of string

if decimal point found, use it as the anchor (center), otherwise use the end of string as anchor

then scan left with multiplier by 10 for every iteration

and scan right of anchor (if anchor is not end of string) and multiplier by 0.1 every iteration

then account for the sign