atof problem..

[ekobudy]

Hello guys,
i'm new in C..

i have code like this :

#include <stdio.h>
#include <stdlib.h>

char *inputstr;
double atofret;

void main() {
inputstr="1234567890.1234";
atofret=atof(inputstr);
printf(" Converted to : %14.4f \n",atofret);
}

the result when i run the program is :
Converted to : 12345678901234.1230


The question is ?
why the string input and the atof output aren't same 1234567890.1234 ?

and if there any way to make it same input and output.. can somebody tell me ?

Thank you.

regards,

Ekow

 

[Salem]

> void main() {
main returns an int, as in

int main ( ) {
  // your code
  return 0;
}

> why the string input and the atof output aren't same
floats and doubles are approximations. The best you'll ever get out of a double variable is about 15 decimal digits of precision.
Your example fails at the 14th decimal digit.

> and if there any way to make it same input and output.. can somebody tell me ?
Only by using high precision maths libraries like say

http://www.swox.com/gmp/

is this likely.

--

 

[xwb]

Some C compilers have the concept of a long double. Maybe you could try that.

 

[mseth]

Using MinGW ( gcc 3.3.1) I'm getting long double being 12 bytes, and double as being 8.

 

[jmnagi]

 inputstr="1234567890.1234";
    atofret=atof(inputstr);
    printf(" Converted to : %14.4f \n",atofret);

hi .. although you will get away with this in most compilers, don't do this without allocating memory first. you will seg fault.

also, atoi, atof etc. explicity look for a NULL at the end. so, a better way to do this will be to add a NULL [after finding string length] and then invoking atof.

br:

 

[mseth]

String literals are automatically null-terminated (by the compiler) as far as I know:

http://c2.com/cgi/wiki?NonNullTerminatedString

A string in C is simply an array of characters, with the final character set to the NUL character (ascii/unicode point 0). This null-terminator is required; a string is ill-formed if it isn't there. The string literal token in C/C++ ("string") guarantees this.
const char *str = "foo";

is the same as
const char *str = {'f', 'o', 'o', 0};

 

[ekobudy]

xwb, can you tell me about long double and where can i find these compiler?

 

[jmnagi]

sethmcdoogle <= it is valid only at variable declaration, only at application level [user mode] however.

in this case, declaration and assignment are different instructions. besides, as mentioned in my previous post, we will get away with it in most compilers .. not always the case.

br:

 

[xwb]

Which OS are you using?

As sethmcdoogle mentioned, it is available on gcc (Unix & Windows).

On Windows, it is available on Visual C++ as well.

 

[ekobudy]

Ok, thank you xwb.
I'm working on Solaris9.

thanks, for the information.
I will download gcc first.

Thank you to all of you guys.

regads,

ekow

 

[mingis]

jmnagi:
> don't do this without allocating memory first. you will seg fault.

everything seem to be allocated?

 

[Salem]

jmnagi is wrong.

char *inputstr="1234567890.1234";

Is both always allocated, and always nul-terminated.

Where you can come unstuck is when you try and modify a string constant. Given the above assignment

inputstr[0] = '2';

Most modern combinations of OS and compiler will generate code which will trap when you attempt this assignment.

"string" constants can be placed in read-only memory. If the compiler chooses to, and the OS enforces it, attempts to modify read-only memory will generate an exception.

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