Here is information on the binary format of double (at least ones that follow the IEEE standard)
Printable View
Here is information on the binary format of double (at least ones that follow the IEEE standard)
Ok, I coded a simple representation of an extended Atof:
I don't understand why even declaring my vars as long double, I can't get the precision offered by this type.Code:#include <stdio.h>
#include <string.h>
#include <ctype.h>
#include <stdlib.h>
#include <math.h>
#ifndef MAXSIZE
#define MAXSIZE 1000
#endif
enum bool {TRUE = 1, FALSE = 0};
long double Atof(char* s)
{
enum bool isNegative = FALSE;
long double val, power;
int i, sign, exp;
/* while it's a space, keep iterating */
for(i = 0; isspace(s[i]); i++)
;
/* record the sign */
sign = (s[i] == '-')? -1 : 1;
/* if it is a sign, just ignore it iterating to the next index */
if(s[i] == '-' || s[i] == '+')
i++;
/* get the power and the value for further manipulation */
/* ----------------------------------------------------- */
for(val = 0.0; isdigit(s[i]); i++)
val = 10 * val + (s[i] - '0');
if(s[i] == '.')
i++;
for(power = 1.0; isdigit(s[i]); i++)
{
val = 10 * val + (s[i] - '0');
power *= 10;
}
/* ----------------------------------------------------- */
/* just ignore if 'e' or 'E' was entered */
if(s[i] == 'e' || s[i] == 'E')
{
i++;
/* test if the exponent is negative */
if(s[i] == '-')
isNegative = TRUE;
i++;
/* test if the next char is a digit */
if(isdigit(s[i]))
exp = s[i] - '0';
/* if the exponent is negative, make the exp value negative
* and return the proper value */
if(isNegative)
{
exp = -exp;
return (sign * val / power) * pow(10.0, exp);
}
/* the exponent isn't negative, so don't make the exp value negative */
else {
return (sign * val / power) * pow(10.0, exp);
}
}
/* 'e' or 'E' wasn't entered, just return the value with the adequate sign */
else {
return val * sign / power;
}
}
int main(void)
{
char strnum[MAXSIZE];
long double d;
printf("Please enter a number in scientific notation of the form ");
printf("[number]e-[exponent] or [number]E-[exponent]: ");
fgets(strnum, sizeof(strnum), stdin);
d = Atof(strnum);
printf("The double value in scientific notation is %Lf\n", d);
return 0;
}
Did you try adjusting the precision of your floating point answer? Here:
The number to the left of the decimal point is the width of the display you want, the number to the right is the precision you want to carry the value to. You can use the width to create space on your screen. Try not to go too crazy on the precision though. There is a hard limit on the number of digits supported by floating point types so there will be a point where you don't get any more precise. See <float.h> for details.Code:int main(void)
{
char strnum[MAXSIZE];
long double d;
printf("Please enter a number in scientific notation of the form ");
printf("[number]e-[exponent] or [number]E-[exponent]: ");
fgets(strnum, sizeof(strnum), stdin);
d = Atof(strnum);
printf("The double value in scientific notation is %18.18Lf\n", d);
return 0;
/*my output:
Please enter a number in scientific notation of the form [number]e-[exponent] or
[number]E-[exponent]: 2.33e-6
The double value in scientific notation is 0.000002330000000000
*/
}
Many thanks. I overlooked that.Quote:
Did you try adjusting the precision of your floating point answer?