Convert String user input to a double

Question

I need to know how to convert a user input, which is a string, to a double. like if he writes in the string "23.45", it converts into double 23.45 (without any library functions).

I already got this code for integer, but don't know how to continue with double:

#include <stdio.h>

void main()
{
    char input[100];
    printf("Type a String which will be converted to an Integer: ");
    scanf("%s", input);

    int number = 0;
    int i = 0;

    if (input[i] >= 48 && input[i] <= 57)
    {
        while (input[i] >= '0' && input[i] <= '9')
        {
            number = number * 10;
            number = number + input[i] - '0';
            i++;
        }
        printf("string %s -> number %d \n", input, number);
    }
    else
    {
        printf("Enter a number! \n");
    }
}

Answer 1

There's probably no reason why you'd roll out your own version of this, as strtod in stdlib.h already covers all manner of formats.

Here's a version which covers signed numbers as input and has some hints of where more suitable error handling could be placed:

#include <stdbool.h>

static void halt_and_catch_fire (void);

double strtod_homebrewn (const char* str)
{
  double result = 0;

  // handle signs:  
  bool is_negative = false;
  if(*str == '-')
  {
    is_negative = true;
    str++;
  }
  else if(*str == '+')
  {
    str++;
  }

  // handle the dot position:
  bool is_dot_found = false;
  double multiplier = 0.1;

  // the actual conversion:
  for(const char* s=str; *s!='\0'; s++)
  {
    if(*s >= '0' && *s <= '9') // ctype.h isdigit() would be preferred here
    {
      if(is_dot_found)
      {
        result += (*s - '0') * multiplier;
        multiplier /= 10;
      }
      else
      {
        result *= 10;
        result += *s - '0';
      }
    }
    else if(*s == '.')
    {
      if(is_dot_found) // two dots?
      {
        halt_and_catch_fire(); // replace this with error handling
      }

      is_dot_found = true;
    }
    else if(*s != '\0') // all cases tested, some weird unknown character found
    {
      halt_and_catch_fire(); // replace this with error handling
    }
  }


  if(is_negative)
  {
    result = -result;
  }

  return result;
}

static void halt_and_catch_fire (void)
{
  halt_and_catch_fire();
}

Answer 2

Edit: only use this approach for small number of digits after the decimal point. Read the comments to know why it would fail for a large number of digits.

Since you mentioned without using any library functions, you could do something like this.

float number;
int decimal = 0;
int decimal_found =10;
while(input[i]!='\0')
{
    if((input[i] <='0' || input[i] >='9')&&input[i]!='.' )
    break;

    if(input[i] == '.')
    decimal = 1;

    if(decimal == 1)
    {
        number = number + (input[i] - '0')/decimal_found;
        decimal_found = decimal_found*10;
    }
    else
    {
        number =  number *10;
        number = number + input[i] - '0';
    }
    i++;
}

Simply check a decimal variable to know when decimal has been reached, then use and if else to have separate conditions for the number variable

Answer 3

Edit: As clux pointed out, this fails when the fraction starts with zeroes. Bummer. Anyway, perhaps someone conceives a simple fix? I can only think of adding a "readzeroes()" function and let that run after the dot.

You already have a function to read an int. Simply use that. Pseudo code:

float read_float()
{
    float f = read_int()
    if(next is dot) skipdot else return f;
    float frac = read_int() 
    while (frac>1) frac /= 10
    return f+frac;
}

Answer 4

#include <stdio.h>

void main()
{
    char input[100];
    printf("Type a String which will be converted to a double: ");
    scanf("%s", input);

    double number = 0.0;
    double divider = 1.0;
    int inFraction = 0;
    int i = 0;

    if (input[i] >= 48 && input[i] <= 57)
    {
        inFraction = 0;
        while ((input[i] >= '0' && input[i] <= '9') || input[i] == '.')
        {
            if (input[i] == '.')
            {
                i++;
                inFraction = 1;
                continue;
            }

            number = number * 10.0;
            number = number + input[i] - '0';
            i++;

            if (inFraction) divider *= 10.0;
        }

        number /= divider;

        printf("string %s -> number %g \n", input, number);
    }
    else
    {
        printf("Enter a number! \n");
    }
}