----- Original Message -----
From: Tim Prince <n8tm@xxxxxxx>
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
