Re: problem with precision when reading from file

[Tim Prince <n8tm at aol dot com>]

 On 9/16/2010 5:38 AM, Anna Sidera wrote:
> ----- Original Message -----
> From: Tim Prince<n8tm@aol.com>
> Date: Wednesday, September 15, 2010 6:07 pm
> Subject: Re: problem with precision when reading from file
>
> >   On 9/15/2010 7:37 AM, Anna Sidera wrote:
> > > Hello,
> > >
> > > I wrote the following program:
> > >
> > >
> > >
> > > #include<stdio.h>
> > > #include<math.h>
> > > #include<stdlib.h>
> > > #include<time.h>
> > > #include<string.h>
> > >
> > > int main()
> > > {
> > >
> > > FILE *fp;
> > > fp = fopen("test.c", "r");
> > > if (fp == NULL){
> > >       printf("Error opening input file\n");
> > >       exit(1);
> > > }
> > >
> > > float var1, var2, var3;
> > >
> > > fscanf(fp,"%f",&var1);
> > > fscanf(fp,"%f",&var2);
> > > fscanf(fp,"%f",&var3);
> > >
> > > if (var1!=0.01) printf("var1 = %g\n",var1);
> > > if (var2!=5) printf("var2 = %g\n",var2);
> > > if (var3!=0.6) printf("var3 = %g\n",var3);
> > >
> > >
> > >
> > > Where test.c contains the following:
> > >
> > > 0.01
> > > 5
> > > 0.6
> > >
> > >
> > >
> > > When I run the program I get the following output:
> > >
> > > var1 = 0.01
> > > var3 = 0.6
> > >
> > >
> > >
> > > I expected to get no output. Can you explain what is the problem?
> > >
> > > Thanks,
> > > Anna
> > Did you mean to try
> >
> > if (var1!=0.01f) printf("var1 = %g\n",var1);
> > if (var3!=0.6f) printf("var3 = %g\n",var3);
> >
> > or possibly to convert var1 and var3 back to short char strings
> > and compare those?
> >
> > 0.01 and 0.6 are constants of type double and will have typically
> > 29 more bits precision in their
> >   conversion from decimal to binary representation than your
> > conversions to type float.
> >
> >
> >
> >
> > --
> > Tim Prince
>
> I wrote a program that uses some input parameters. The program should read the input parameters from a file. There are also some input parameters that are arguments to main. Is there a way to read the input parameters so that for example var1*var2 gives a value equal to 0.05? What is the best way to read the input parameters?
>
> If var1 and var3 are float and 0.01 and 0.6 are long then why you mention short char?
>
> Anna
I meant use sprintf() or such to make a string of char to see if it 
comes out to the desired value.  Such a string conversion might be a 
possibility for your product test.  It should be evident that you can't 
make an inexact float value come out identical to a double value 
resulting from a higher precision calculation, but maybe you haven't 
explained what you want to do.  My initial impression was that you might 
have forgotten the distinction between float and double constants.

--
Tim Prince