Converting string, to float (without atof) in C

Question

I'm designing a function that will convert a string into a float. e.g. "45.5" = 45.5

I have this so far. But it doesn't seem to work. Keep in mind, we cannot use any C library functions like atoi, atof or even pow for that matter.

int str2float( char *s )
{
    int num = 0;
    int dec = 0;
    double i = 1.0;
    int ten = 1;
    /***** ADD YOUR CODE HERE *****/

    for(; *s != '\0'; s++)
    {
        if (*s == '.'){
            for(; *s != '\0'; s++){
                dec = (dec * CONT) + (*s - '0');
                i++;
            }
        }else{
            num = (num * CONT) + (*s - '0');
        }

    }
    for(;i!=0;i--){
        ten *= 10;
    }
    dec = dec / (ten);
    printf("%d", dec);
    num += dec;
    return num;  
}

Answer 1

Here is my try:

float stof(const char* s){
  float rez = 0, fact = 1;
  if (*s == '-'){
    s++;
    fact = -1;
  };
  for (int point_seen = 0; *s; s++){
    if (*s == '.'){
      point_seen = 1; 
      continue;
    };
    int d = *s - '0';
    if (d >= 0 && d <= 9){
      if (point_seen) fact /= 10.0f;
      rez = rez * 10.0f + (float)d;
    };
  };
  return rez * fact;
};

Answer 2

One potential issue is that s is incremented by the outer loop before checking that it isn't pointing to the NULL terminator.

for(; *s != '\0'; s++)
{
        ...
        for(; *s != '\0'; s++){
        ...
        }
        // inner loop is done now since we have *s=='\0' ...
    ...
    // ... but now we're going to increment s again at the end of the outer loop!
}

You need to exit both the inner and outer loop immediately after the NULL terminator is spotted.