Convert a string to float without using inbuilt function in c

Question

I don't know how to convert a string value to double or float. Here is the sample string I am trying to convert: "3.45667". I am struggling with the handling of the dot.

I tried this but it is not working:

#include<stdio.h>

int main()
{
    char string_array[] = { "3.45667" };
    float float_array[20], index = 0;

    for(index = 0 ; index < 7 ; index++)
    {
        if(string_array[index] == '.')
        {
            printf("dot");
            // here, how to add dot to the float_array?
        } else
        {
            float_array = (float)(string_array[index] - '0');
        }
    }
    return 0;
}

How to add a decimal point in the above code? I think this could be done with exponent form but I don't know how to use exponent form.

Nim · Accepted Answer · 2011-08-17 13:25:03Z

2

Here is an algorithm..

Find the .
Get a substring to that point and convert to an integer (iterate through the characters, convert to int - should be a simple problem)
Do the same thing with everything after, also keep track of number of digits in the "fractional" component.
Divide the "fractional" component by 10 to the power of the number of digits, and then add that to the integral component.

EDIT: if not using a C function, and given the c++ tag, I'd go with..

std::istringstream foo("3.141");
double v;
foo >> v;

No C function in sight! ;)

answered Aug 17, 2011 at 13:25

Nim

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2 Comments

the wolf Over a year ago

That is C++. The OP was for C.

Nim Over a year ago

@carrot-top, obviously! The original question was tagged c++!

Ingo · Accepted Answer · 2011-08-17 23:24:03Z

0

The following program just counts how many digits are after the decimal point and scales the result in the end, instead of each digit.

float result = 0.0;
float scale  = 0.0;

char *s = "3.14567":

while (*s) {
     if (*s >= '0' && *s <= '9') {
        result = 10.0 * result + (float)(*s - '0');
        scale *= 10.0;
     }
     else if (*s == '.') scale = 1.0;
     else  { /* ERROR HANDLING */ }
     s++;
}
result = result / (scale == 0.0 ? 1.0 : scale)

edited Aug 17, 2011 at 23:24

answered Aug 17, 2011 at 13:57

Ingo

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2 Comments

user688635 Over a year ago

This does not work. There is an infinite loop since the pointer *s is never incremented. Also, if you fix that, the digits to the right of the decimal are not handled.

Ingo Over a year ago

@Colt, s is incremented in every loop in the last line of the while block. The error handling is supposed to be a block, of course, the indentation makes it clear that s++ is not meant to be part of it. But if you prefer, I'll add a semicolon to make it fit for copy-and-paste, which is, apparently, the only thing that counts. Your other statement is just plain false in the light of this.

Patrick87 · Accepted Answer · 2011-08-17 13:30:12Z

0

Use the position of the decimal point in the string to scale the digits by the appropriate factor of 10 when you go to sum them... so the nth item to the left of the decimal is scaled by 10^(n-1) and the mth item to the right is scaled by (0.1)^m. So...

 for each char in thestr do
    if it's a digit then
       Add the digit to the list
    else then
       dp = position - 1

 sum = 0
 for each digit in list
    sum = sum + digit * pow(10, dp - position)

 return sum

answered Aug 17, 2011 at 13:30

Patrick87

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Comments

duedl0r · Accepted Answer · 2011-08-17 14:17:41Z

0

int num = 0;

// parse the numbers before the dot
while(*string != '.')
  num += 10*num + *string++ - '0';

string++;
float frac = 0.0f;
float div = 1.0f;

// parse the numbers after the dot
while(*string != '\0') {
  frac += (*string++ - '0')/div;
  div /= 10.0f;
}

float res = num + frac;

or something like this.. most probably some bugs in it..

edited Aug 17, 2011 at 14:17

answered Aug 17, 2011 at 13:34

duedl0r

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3 Comments

duedl0r Over a year ago

hm? I don't know what you mean.. I just wrote some minimal code, because I'm too lazy to write an algorithm in words.. If you need some error checks, you can do that on your own :)

Taras Mazepa Over a year ago

oh, come on, you write some code without this boring checks to the low skilled programmer - of course for this exist stackoverflow. you must explain something not solve some easy exercises for practice

duedl0r Over a year ago

I still don't get your question :)

dawg · Accepted Answer · 2011-08-17 16:53:50Z

This works in a pinch:

#include <ctype.h>

double _atof(char *s)
{
        double a = 0.0;
        int e = 0;
        int c;
        while ((c = *s++) != '\0' && isdigit(c)) {
                a = a*10.0 + (c - '0');
        }
        if (c == '.') {
                while ((c = *s++) != '\0' && isdigit(c)) {
                        a = a*10.0 + (c - '0');
                        e = e-1;
                }
        }
        if (c == 'e' || c == 'E') {
                int sign = 1;
                int i = 0;
                c = *s++;
                if (c == '+')
                        c = *s++;
                else if (c == '-') {
                        c = *s++;
                        sign = -1;
                }
                while (isdigit(c)) {
                        i = i*10 + (c - '0');
                        c = *s++;
                }
                e += i*sign;
        }
        while (e > 0) {
                a *= 10.0;
                e--;
        }
        while (e < 0) {
                a *= 0.1;
                e++;
        }
        return a;
}

You will have a double and then you will need to convert or cast to a float.

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