Problem with printf formats

[Guenther Sohler]

I have following code:

int main(void)
{
printf("%.3lf\n",-2158470*0.001);
}

it prints

-2158.470

How can i disable '0' at the end, if they are not needed ?

Is there a possibility just to write %lf and just print as many digits
as needed to represent the number ?

 

[Chuck F.]

>
I have following code:

int main(void)
{
printf("%.3lf\n",-2158470*0.001);
}

it prints

-2158.470

How can i disable '0' at the end, if they are not needed ?

Is there a possibility just to write %lf and just print as many
digits as needed to represent the number ?

You are asking the wrong question. The trailing zero is needed to
represent the number. Otherwise how could you distinguish it from:

printf("%.2f\n", -2158471 * 0.001)

(always assuming that -2158471 does not create an overflow)

Your question, I believe, is "how to suppress trailing zeroes". A
little thought and an intermediary buffer should give you a
suitable answer for that.

 

[usenet]

Guenther Sohle said:

(...)
(...)
it prints

-2158.470

How can i disable '0' at the end, if they are not needed ? Is there a
possibility just to write %lf and just print as many digits as needed to
represent the number ?

The problem is not in the printf() function, but in the way floating point
numbers are represented by your computer. Due to rounding errors, floating
point numbers hardly ever have the 'exact' value you would expect. Search
google for 'floating point' for some details, or read this wikipedia page:
http://en.wikipedia.org/wiki/Floating_point.

It is up to you to make the choice how many digits are significant in a
number, so that is why you should tell printf() about this.

 

[Jirka Klaue]

Guenther Sohler:

int main(void)
{
printf("%.3lf\n",-2158470*0.001);
}

it prints

-2158.470

How can i disable '0' at the end, if they are not needed ?

Is there a possibility just to write %lf and just print as many digits
as needed to represent the number ?

The l in %lf is superfluous, just use %f.
To avoid the trailing zeros try %g.

Jirka

 

[Guenther Sohler]

Guenther Sohler:


The l in %lf is superfluous, just use %f. Thank you for the info.
To avoid the trailing zeros try %g.

I also tried %g:
int main(void)
{
printf("%g\n",-2158475*0.001);
}

which results in
-2158.47

one digit is missed! this is a killing issue for me!

how can i have the proper value displayed in all cases ?

 

[Guenther Sohler]

Guenther Sohle said:

The problem is not in the printf() function, but in the way floating point
numbers are represented by your computer. Due to rounding errors, floating
point numbers hardly ever have the 'exact' value you would expect. Search
google for 'floating point' for some details, or read this wikipedia page:
http://en.wikipedia.org/wiki/Floating_point.

It is up to you to make the choice how many digits are significant in a
number, so that is why you should tell printf() about this.

I know, but the accuracy of double should still good enough to fulfill my
purpose!

 

[Guenther Sohler]

Your question, I believe, is "how to suppress trailing zeroes". A
little thought and an intermediary buffer should give you a
suitable answer for that.

Thank you for the information. Yes, I also keep this in mind,
but i use %f very often in my code ( > 100) and to suitable
handle all them, I would have to write my own printf.
I dont like to do this unless I am really sure there is no other way

 

[Jirka Klaue]

Guenther Sohler:

I also tried %g:
int main(void)
{
printf("%g\n",-2158475*0.001);
}

which results in
-2158.47

one digit is missed! this is a killing issue for me!

Well, then it's time for you to read the appropriate part
of the standard.

ISO/IEC 9899:TC2
7.19.6.1#8
g,G A double argument representing a floating-point number is converted
in style f or e (or in style F or E in the case of a G conversion
specifier), depending on the value converted and the precision.
Let P equal the precision if nonzero, 6 if the precision is omitted,
or 1 if the precision is zero. Then, if a conversion with style E
would have an exponent of X:
- if P > X ≥ -4, the conversion is with style f (or F) and
precision P - (X + 1).
- otherwise, the conversion is with style e (or E) and precision P - 1.
Finally, unless the # flag is used, any trailing zeros are removed from
the fractional portion of the result and the decimal-point character is
removed if there is no fractional portion remaining.

So try %.9g or something like that.

how can i have the proper value displayed in all cases ?

This could be difficult depending on the meaning of proper.

Jirka

 

[Dik T. Winter]

> int main(void)
> {
> printf("%g\n",-2158475*0.001);
> }
>
> which results in
> -2158.47
>
> one digit is missed! this is a killing issue for me!
>
> how can i have the proper value displayed in all cases ?

But, what *is* the proper value? In binary it is:
-10000110110.01110011001100110011...
and that is what is stored on the computer, most decimal numbers are
not exactly representable in binary. So what will be stored is a
rounded value, and the value in memory is:
-2158.47499999999990905052982270717620849609375
so what should the system do?

 

[Old Wolf]

Guenther Sohler:

The l in %lf is superfluous, just use %f.

It is superfluous in C99, and is an error in C89. But most
C89 compilers allow it anyway.

 

[Tatu Portin]

I have following code:

int main(void)
{
printf("%.3lf\n",-2158470*0.001);
}

it prints

-2158.470

How can i disable '0' at the end, if they are not needed ?

Is there a possibility just to write %lf and just print as many
digits as needed to represent the number ?

Question "How can I disable '0' at the end" indicates somekind of
misconception of floating point numbers.

Count of digits in a number indicates the accuracy of the number. So,
if you just remove zero from a number, you decrease accuracy of the
number.

For example:
-2158.5 could be result of rounding -2158.470, but it would be wrong
to add zeros to -2158.5 because the accurate amount was -2158.470.

The significance of zeros in indicating accuracy can be only perceived
when handling numbers on right side of decimal point.
For example:
-2000 could be result of rounding -2158 or it could be the exact
amount. You cannot tell if -2000 has one, two, three or four
significant digits, i.e. if -2000 is accurate to first, second, third
or fourth digit.
This is why scientific notation is used to represent quantities.
E.g. -2.000e3 is different from -2e3 in accuracy.


--
It's bit too late to be clear, but I hope I made my point through.

C faq: http://www.eskimo.com/~scs/C-faq/top.html
Reference: http://www.acm.uiuc.edu/webmonkeys/book/c_guide/
Coding standards: http://www.psgd.org/paul/docs/cstyle/cstyle.htm

 

[Andrey Tarasevich]

...
I know, but the accuracy of double should still good enough to fulfill my
purpose!
...

What exactly do you mean by "good enough" and "your purpose" in this case? Most
of the time floating-point numbers that have innocent-looking decimal
representations have infinitely long [periodic] representations in traditional
binary positional notation. For example, both '0.001' and '2158.47' are such
numbers. It is impossible to represent them precisely in 'double', which
probably means that 'double's precision (or precision of any type of same
structure, no matter how large) is definitely not "good enough" (assuming that I
understood your "good enough" correctly).

 

[Gordon Burditt]

I know, but the accuracy of double should still good enough to fulfill my

What exactly do you mean by "good enough" and "your purpose" in this case?

(I'm not the original poster).

It is difficult and very expensive to measure most any physical
quantity to more significant digits than are present in a double.
(15 in a typical IEEE float implementation). Exceptions may include
currency and time.

Gordon L. Burditt

 

[Simon Biber]

I have following code:

int main(void)
{
printf("%.3lf\n",-2158470*0.001);
}

it prints

-2158.470

How can i disable '0' at the end, if they are not needed ?

Is there a possibility just to write %lf and just print as many digits
as needed to represent the number ?

I can write a function to do it.

#include <stdio.h>
#include <string.h>

char *sprintd(char *buf, size_t buf_size, double val, int precision)
{
/* write value into string */
snprintf(buf, buf_size, "%.*f", precision, val);

/* find decimal point */
char *p = strchr(buf, '.'), *q;

/* remove trailing zeros */
if(p) for(q = strchr(p, 0) - 1; *q == '0'; --q) *q = 0;

/* remove trailing decimal point */
if(p && p[1] == 0) p[0] = 0;

return buf;
}

int main(void)
{
char buf[100];
printf("%s\n", sprintd(buf, sizeof buf, 355.0 / 113.0, 15));
return 0;
}

 

[vishnu]

write %.2lf inn place of %.3lf

 

[Keith Thompson]

write %.2lf inn place of %.3lf

What?

Read <http://cfaj.freeshell.org/google/>.

 

[Joe Wright]

I know, but the accuracy of double should still good enough to fulfill my
purpose!

You are perhaps confusing accuracy and precision. A double will always
be precise to 53 bits or 16 digits or whatever. Its value is often less
accurate than that, depending on how it was contrived.