Re: float

Re: float

[Bryan Christ]

try

float i=1.34F

On Fri, 2010-01-08 at 16:30 +0000, phoenixxz wrote:
> float i=1.34;
> printf("%f",i);
> 
> 
> ->> 0
> why?
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float

[phoenixxz]

float i=1.34;
printf("%f",i);


->> 0
why?

Re: float

[Nanakos Chrysostomos]

On 8/1/2010 18:30, phoenixxz wrote:
> float i=1.34;
> printf("%f",i);
>
>
> ->> 0
> why?
> --
> To unsubscribe from this list: send the line "unsubscribe linux-c-programming" in
> the body of a message to majordomo@vger.kernel.org
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>   
printf("%lf",i);

Re: float

[Fawad Lateef]

Hello,


On Fri, Jan 8, 2010 at 4:30 PM, phoenixxz <vim.xxz@gmail.com> wrote:
>
> float i=1.34;
> printf("%f",i);
>
>
> ->> 0
> why?
> --


This should be fine. You might be doing some mistake. Can you give
some more details about what you want to do ?

--
Fawad Lateef

Re: float

[Fawad Lateef]

On Fri, Jan 8, 2010 at 4:42 AM, Bryan Christ <bryan.christ@hp.com> wrote:
> try
>
> float i=1.34F

On linux (gcc compiler) it will work without 'F' too.

>
> On Fri, 2010-01-08 at 16:30 +0000, phoenixxz wrote:
>> float i=1.34;
>> printf("%f",i);
>>
>>
>> ->> 0
>> why?
>> --
>> To unsubscribe from this list: send the line "unsubscribe linux-c-programming" in
>> the body of a message to majordomo@vger.kernel.org
>> More majordomo info at  http://vger.kernel.org/majordomo-info.html
>
>
> --
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>



-- 
Fawad Lateef
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Re: float

[Nanakos Chrysostomos]

On 8/1/2010 18:30, phoenixxz wrote:
> float i=1.34;
> printf("%f",i);
>
>
> ->> 0
> why?
> --
> To unsubscribe from this list: send the line "unsubscribe linux-c-programming" in
> the body of a message to majordomo@vger.kernel.org
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>   
printf("%lf",i);

Re: float

[Michal Nazarewicz]

> On 8/1/2010 18:30, phoenixxz wrote:
>> float i=1.34;
>> printf("%f",i);
>>
>> ->> 0
>> why?

Nanakos Chrysostomos <linuxcprog@wired-net.gr> writes:
> printf("%lf",i);

%lf is actually invalid format (though glibc will gladly treat it the
 same way it treats %f).  When passing a float value to a function with
 variable number of arguments it is promoted to double.

-- 
Best regards,                                         _     _
 .o. | Liege of Serenly Enlightened Majesty of      o' \,=./ `o
 ..o | Computer Science,  Michal "mina86" Nazarewicz   (o o)
 ooo +--<mina86*tlen.pl>--<jid:mina86*jabber.org>--ooO--(_)--Ooo--

Re: float

[Nanakos Chrysostomos]

On 8/1/2010 19:11, Michal Nazarewicz wrote:
>> On 8/1/2010 18:30, phoenixxz wrote:
>>     
>>> float i=1.34;
>>> printf("%f",i);
>>>
>>> ->> 0
>>> why?
>>>       
> Nanakos Chrysostomos <linuxcprog@wired-net.gr> writes:
>   
>> printf("%lf",i);
>>     
> %lf is actually invalid format (though glibc will gladly treat it the
>  same way it treats %f).  When passing a float value to a function with
>  variable number of arguments it is promoted to double.
>
>   
It could then be:

printf("%lf", (double)i);

Re: float

[Nanakos Chrysostomos]

On 8/1/2010 19:11, Michal Nazarewicz wrote:
>> On 8/1/2010 18:30, phoenixxz wrote:
>>     
>>> float i=1.34;
>>> printf("%f",i);
>>>
>>> ->> 0
>>> why?
>>>       
> Nanakos Chrysostomos <linuxcprog@wired-net.gr> writes:
>   
>> printf("%lf",i);
>>     
> %lf is actually invalid format (though glibc will gladly treat it the
>  same way it treats %f).  When passing a float value to a function with
>  variable number of arguments it is promoted to double.
>
>   
Sorry for the previous one...
it could be then.

printf("%f",(double)i);

Re: float

[Michal Nazarewicz]

> On 8/1/2010 19:11, Michal Nazarewicz wrote:
>> %lf is actually invalid format (though glibc will gladly treat it the
>>  same way it treats %f).  When passing a float value to a function with
>>  variable number of arguments it is promoted to double.

Nanakos Chrysostomos <linuxcprog@wired-net.gr> writes:
> it could be then.
>
> printf("%f",(double)i);

Expect explicit cast is not required here as (in case of "..."
arguments) it is done implicitly as I described.

-- 
Best regards,                                         _     _
 .o. | Liege of Serenly Enlightened Majesty of      o' \,=./ `o
 ..o | Computer Science,  Michal "mina86" Nazarewicz   (o o)
 ooo +--<mina86*tlen.pl>--<jid:mina86*jabber.org>--ooO--(_)--Ooo--

Re: float

[andrej.gelenberg]

Hi,

work for me.

cat >test.c <<END
#include <stdio.h>
int main(){
 float i=1.34;
 printf("%f\n",i);
 return 0;
}
END
gcc -o test test.c && ./test

output:
1.340000

phoenixxz writes:

> float i=1.34;
> printf("%f",i);
> 
> 
> ->> 0
> why?
> --
> To unsubscribe from this list: send the line "unsubscribe linux-c-programming" in
> the body of a message to majordomo@vger.kernel.org
> More majordomo info at  http://vger.kernel.org/majordomo-info.html

Re: float

[Glynn Clements]

phoenixxz wrote:

> float i=1.34;
> printf("%f",i);
> 
> 
> ->> 0
> why?

You need to include <stdio.h> and write a main() function ;)

Beyond that: stdout will be line-buffered (if it's associated with a
terminal, block-buffered otherwise), so you should add a "\n" to the
end of the format string.

The lack of line buffering might have an effect if the program
terminates abnormally (exit() or returning from main() should cause
buffered streams to be flushed).

A working example:

	$ cat test.c
	#include <stdio.h>
	int main(void)
	{
	    float i=1.34;
	    printf("%f\n",i);
	    return 0;
	}
	$ gcc test.c
	$ ./a.out
	1.340000

-- 
Glynn Clements <glynn@gclements.plus.com>

Re: float

[Thomas Weber]

Am 11.01.2011 09:34, schrieb ratheesh k:
> On Tue, Jan 11, 2011 at 1:31 PM, Thomas Weber
> <thomas.weber.linux@googlemail.com> wrote:
>> Okay, now I got it.
>>
>> Look for IEEE754.
>>
>> 0.1 is in double precision 0.10000000149011612
>>
>> Thomas
>>
> 
> Is it possible to print the correct "printf" statement by typecasting
> some of operations or  data types. I tried with no success.
> 
> -Ratheesh

Try something with precision and typecast float to double

printf("f:%.10f is 1.0 \n",(double) f);


Thomas

> 
> 
>> Am 11.01.2011 08:55, schrieb Uriel Corfa:
>>> I'd say : rounding errors. Floating point goes in mysterious ways.
>>>
>>> On Tue, Jan 11, 2011 at 8:38 AM, ratheesh k <ratheesh.ksz@gmail.com> wrote:
>>>> I could not understand  why it getting printed like this. Could any
>>>> body tell me.
>>>>
>>>> #include <stdio.h>
>>>>
>>>> int main()
>>>> {
>>>>        float f=0.0f;
>>>>        int i;
>>>>
>>>>        for(i=0;i<10;i++)
>>>>                f = f + 0.1f;
>>>>
>>>>        if(f == 1.0f)
>>>>                printf("f is 1.0 \n");
>>>>        else
>>>>                printf("f is NOT 1.0\n");
>>>>
>>>>        return 0;
>>>> }
>>>
>>

Re: float

[Andrej Gelenberg]

Hi,

you can not check if 2 float are equal, you should check if it in the range:

On 01/11/2011 08:38 AM, ratheesh k wrote:
> I could not understand  why it getting printed like this. Could any
> body tell me.
>
> #include<stdio.h>
>

#define ABS(x) ((x) < 0 ? -(x) : (x))
#define MAX(a, b) ((a) > (b) ? (a) : (b))
const float precision = 0.001f;

> int main()
> {
>          float f=0.0f;
>          int i;
>
>          for(i=0;i<10;i++)
>                  f = f + 0.1f;
>
            if ( ABS(f - 1.0f) < precision )
>                  printf("f is 1.0 \n");
>          else
>                  printf("f is NOT 1.0\n");
>
>          return 0;
> }

Here more to that topic: 
https://www.securecoding.cert.org/confluence/display/seccode/FLP35-C.+Take+granularity+into+account+when+comparing+floating+point+values

Here more about c pitfalls: 
https://www.securecoding.cert.org/confluence/display/seccode/CERT+C+Secure+Coding+Standard

Regards,
Andrej Gelenberg

Re: float

[ratheesh k]

On Tue, Jan 11, 2011 at 1:31 PM, Thomas Weber
<thomas.weber.linux@googlemail.com> wrote:
> Okay, now I got it.
>
> Look for IEEE754.
>
> 0.1 is in double precision 0.10000000149011612
>
> Thomas
>

Is it possible to print the correct "printf" statement by typecasting
some of operations or  data types. I tried with no success.

-Ratheesh


> Am 11.01.2011 08:55, schrieb Uriel Corfa:
>> I'd say : rounding errors. Floating point goes in mysterious ways.
>>
>> On Tue, Jan 11, 2011 at 8:38 AM, ratheesh k <ratheesh.ksz@gmail.com> wrote:
>>> I could not understand  why it getting printed like this. Could any
>>> body tell me.
>>>
>>> #include <stdio.h>
>>>
>>> int main()
>>> {
>>>        float f=0.0f;
>>>        int i;
>>>
>>>        for(i=0;i<10;i++)
>>>                f = f + 0.1f;
>>>
>>>        if(f == 1.0f)
>>>                printf("f is 1.0 \n");
>>>        else
>>>                printf("f is NOT 1.0\n");
>>>
>>>        return 0;
>>> }
>>
>
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Re: float

[Thomas Weber]

Am 11.01.2011 08:55, schrieb Uriel Corfa:
> I'd say : rounding errors. Floating point goes in mysterious ways.
> 
> On Tue, Jan 11, 2011 at 8:38 AM, ratheesh k <ratheesh.ksz@gmail.com> wrote:
>> I could not understand  why it getting printed like this. Could any
>> body tell me.
>>
>> #include <stdio.h>
>>
>> int main()
>> {
>>        float f=0.0f;
>>        int i;
>>
>>        for(i=0;i<10;i++)
>>                f = f + 0.1f;
>>
>>        if(f == 1.0f)
>>                printf("f is 1.0 \n");
>>        else
>>                printf("f is NOT 1.0\n");
>>
>>        return 0;
>> }
> 

Okay, now I got it.

Look for IEEE754.

0.1 is in double precision 0.10000000149011612

Thomas

Re: float

[Amit Virdi]

In floating point comparisons, test for range rather testing for equality. Something like:

 int compare_float(float f1, float f2)
 {
  float precision = 0.00001;
  if (((f1 - precision) < f2) && 
      ((f1 + precision) > f2))
   {
    return 1;
   }
  else
   {
    return 0;
   }
 }

This is because, floating point numbers are stored in binary format and rounds off in strange ways.

On 01/11/2011 01:16 PM, Thomas Weber wrote:
> Am 11.01.2011 08:38, schrieb ratheesh k:
>> I could not understand  why it getting printed like this. Could any
>> body tell me.
>>
>> #include <stdio.h>
>>
>> int main()
>> {
>>         float f=0.0f;
>>         int i;
>>
>>         for(i=0;i<10;i++)
> missing braces here?
>>                 f = f + 0.1f;
>>
>>         if(f == 1.0f)
>>                 printf("f is 1.0 \n");
>>         else
>>                 printf("f is NOT 1.0\n");
>>
>>         return 0;
>> }
>> --
> 
> 
> Thomas
> 
> --
> To unsubscribe from this list: send the line "unsubscribe linux-c-programming" in
> the body of a message to majordomo@vger.kernel.org
> More majordomo info at  http://vger.kernel.org/majordomo-info.html
> 

-- 
~Amit Virdi

Re: float

[Uriel Corfa]

I'd say : rounding errors. Floating point goes in mysterious ways.

On Tue, Jan 11, 2011 at 8:38 AM, ratheesh k <ratheesh.ksz@gmail.com> wrote:
> I could not understand  why it getting printed like this. Could any
> body tell me.
>
> #include <stdio.h>
>
> int main()
> {
>        float f=0.0f;
>        int i;
>
>        for(i=0;i<10;i++)
>                f = f + 0.1f;
>
>        if(f == 1.0f)
>                printf("f is 1.0 \n");
>        else
>                printf("f is NOT 1.0\n");
>
>        return 0;
> }

-- 
Uriel Corfa
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Re: float

[Thomas Weber]

Am 11.01.2011 08:38, schrieb ratheesh k:
> I could not understand  why it getting printed like this. Could any
> body tell me.
> 
> #include <stdio.h>
> 
> int main()
> {
>         float f=0.0f;
>         int i;
> 
>         for(i=0;i<10;i++)
missing braces here?
>                 f = f + 0.1f;
> 
>         if(f == 1.0f)
>                 printf("f is 1.0 \n");
>         else
>                 printf("f is NOT 1.0\n");
> 
>         return 0;
> }
> --


Thomas

float

[ratheesh k]

I could not understand  why it getting printed like this. Could any
body tell me.

#include <stdio.h>

int main()
{
        float f=0.0f;
        int i;

        for(i=0;i<10;i++)
                f = f + 0.1f;

        if(f == 1.0f)
                printf("f is 1.0 \n");
        else
                printf("f is NOT 1.0\n");

        return 0;
}