Convert a String to Float

Code

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

float custom_atof(const char *str) {
    // Your logic here
    float num1=0,num2=0;
    int i,k,c=0,j=0,sign=1;
    float t=1.0f;
    if(str[0]=='-')
    {
    i=1;
     sign=-1;
    }
    else if(str[0]=='+')
    {
        i=1;
        sign=1;
    }
    else 
    i=0;
    
    for(;str[i];i++)
    {
        if(str[i]>='0' && str[i]<='9')
        num1=num1*10+(str[i]-'0');
        else if (str[i]=='.')
        {
             j=i;
            break;
        }
        else
        break;
    }
    for(j=i+1;str[j];j++)
    {
        if(str[j]>='0'&& str[j]<='9')
        {
            num2=num2*10+(str[j]-'0');
            c++;
        }
        else
        break;
    }
    for(k=0;k<c;k++)
    t=t*10.0f;
    num2=num2/t;
     

    return sign * (num1+num2);
}

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

float num1=0,num2=0;
    int i,k,c=0,j=0,sign=1;
    float t=1.0f;
    if(str[0]=='-')
    {
    i=1;
     sign=-1;
    }
    else if(str[0]=='+')
    {
        i=1;
        sign=1;
    }
    else 
    i=0;
    
    for(;str[i];i++)
    {
        if(str[i]>='0' && str[i]<='9')
        num1=num1*10+(str[i]-'0');
        else if (str[i]=='.')
        {
             j=i;
            break;
        }
        else
        break;
    }
    for(j=i+1;str[j];j++)
    {
        if(str[j]>='0'&& str[j]<='9')
        {
            num2=num2*10+(str[j]-'0');
            c++;
        }
        else
        break;
    }
    for(k=0;k<c;k++)
    t=t*10.0f;
    num2=num2/t;
     

    return sign * (num1+num2);