Hi,
I want to represent a double in char* as accurate as possible, but as
short as possible:
double char*
10000 1E5
1000.00123 1000.00123
....
How can I archive this?
%.10f prints too much, and %.10g is too inaccurate.
--
-Gernot
int main(int argc, char** argv) {printf
("%silto%c%cf%c gl%ssic%ccom%c" , "ma", 58, 'g', 64, "ba", 46, 10);}
_______________ _______________ __________
Looking for a good game? Do it yourself!
GLBasic - you can do www.GLBasic.com 14  7978 
"Gernot Frisch" <Me@Privacy.net > wrote in message
news:2s******** *****@uni-berlin.de... I want to represent a double in char* as accurate as possible, but as
short as possible:
This is the intent of the %g formatter in printf.
double char*
10000 1E5
1000.00123 1000.00123
...
How can I archive this?
%.10f prints too much, and %.10g is too inaccurate.
How is %g "too inaccurate", could you provide an example ?
The accuracy of a double, in IEEE format, is limited to about 15 decimal
digits. So a %.15g should do what you asked for.
Regards,
Ivan
-- http://ivan.vecerina.com/contact/?subject=NG_POST <- email contact form
"Ivan Vecerina" <NO************ *************** *******@vecerin a.com>
schrieb im Newsbeitrag news:ck******** **@newshispeed. ch... "Gernot Frisch" <Me@Privacy.net > wrote in message
news:2s******** *****@uni-berlin.de... I want to represent a double in char* as accurate as possible, but
as short as possible:
This is the intent of the %g formatter in printf.
double char*
10000 1E5
1000.00123 1000.00123
...
How can I archive this?
%.10f prints too much, and %.10g is too inaccurate.
How is %g "too inaccurate", could you provide an example ?
The accuracy of a double, in IEEE format, is limited to about 15
decimal
digits. So a %.15g should do what you asked for.
double f = 10000.000000012 3567;
printf("%20.15f - %.15g\n", f, f);
// 10000.000000012 356000 - 10000.000000012 4
Gernot Frisch wrote:
"Ivan Vecerina" <NO************ *************** *******@vecerin a.com>
schrieb im Newsbeitrag news:ck******** **@newshispeed. ch... "Gernot Frisch" <Me@Privacy.net > wrote in message
news:2s******** *****@uni-berlin.de... I want to represent a double in char* as accurate as possible, but
as short as possible:
This is the intent of the %g formatter in printf.
double char*
10000 1E5
1000.00123 1000.00123
...
How can I archive this?
%.10f prints too much, and %.10g is too inaccurate.
How is %g "too inaccurate", could you provide an example ?
The accuracy of a double, in IEEE format, is limited to about 15
decimal
digits. So a %.15g should do what you asked for.
double f = 10000.000000012 3567;
cout the digits
10000.000000012 3567
00000 0000111111111 (Note: 1 reads: 14
12345 6789012345678 4
So everything to the right of
......123
is most likely noise anyway (assuming a standard
precission of 15 digits)
Note: 15 digits does *not* mean: 15 digits after comma.
It means 15 digits total!
--
Karl Heinz Buchegger kb******@gascad .at
Gernot Frisch wrote:
"Ivan Vecerina" <NO************ *************** *******@vecerin a.com>
schrieb im Newsbeitrag news:ck******** **@newshispeed. ch... "Gernot Frisch" <Me@Privacy.net > wrote in message
news:2s******** *****@uni-berlin.de... I want to represent a double in char* as accurate as possible, but
as short as possible:
This is the intent of the %g formatter in printf.
double char*
10000 1E5
1000.00123 1000.00123
...
How can I archive this?
%.10f prints too much, and %.10g is too inaccurate.
How is %g "too inaccurate", could you provide an example ?
The accuracy of a double, in IEEE format, is limited to about 15
decimal
digits. So a %.15g should do what you asked for.
double f = 10000.000000012 3567;
cout the digits
10000.000000012 3567
00000 0000111111111 (Note: 1 reads: 14
12345 6789012345678 4
So everything to the right of
.......000123
is most likely noise anyway (assuming a standard
precission of 15 digits)
Note: 15 digits does *not* mean: 15 digits after comma.
It means 15 digits total!
--
Karl Heinz Buchegger kb******@gascad .at
> cout the digits
10000.000000012 3567
00000 0000111111111 (Note: 1 reads: 14
12345 6789012345678 4
So everything to the right of
.......000123
is most likely noise anyway (assuming a standard
precission of 15 digits)
Note: 15 digits does *not* mean: 15 digits after comma.
It means 15 digits total!
So, why is %.15f printing it correctly? Coincidence?
Gernot Frisch wrote: cout the digits
10000.000000012 3567
00000 0000111111111 (Note: 1 reads: 14
12345 6789012345678 4
So everything to the right of
.......000123
is most likely noise anyway (assuming a standard
precission of 15 digits)
Note: 15 digits does *not* mean: 15 digits after comma.
It means 15 digits total!
So, why is %.15f printing it correctly? Coincidence?
15 is average. It depends on the number itself how many digits you get
exactly. BTW by assining 10000.000000012 3567 and getting
10000.000000012 356000 one cannot talk from correctly. The last
digit you assigned was 7, printf brought back a 0 (and didn't round
the previous 6 to 7).
Might be interesting for you: http://www.petebecker.com/js200006.html http://docs.sun.com/source/806-3568/ncg_goldberg.html
--
Karl Heinz Buchegger kb******@gascad .at
Gernot Frisch wrote: cout the digits
10000.0000000 123567
00000 0000111111111 (Note: 1 reads: 14
12345 6789012345678 4
So everything to the right of
.......000123
is most likely noise anyway (assuming a standard
precission of 15 digits)
Note: 15 digits does *not* mean: 15 digits after comma.
It means 15 digits total!
So, why is %.15f printing it correctly? Coincidence?
Absolutely.
Change the number (to be, e.g., 10000.000000012 3333) and see if
you get the same "precise" output. I got two different results
on Linux (g++ v 3.2.2) and IRIX (MIPSpro 7.3). Both had garbage
after the first 16 digits. sizeof(double) is 8 on both systems.
Also, change the leading 1 to, say, leading 9, and you will get
even "less" accuracy -- the garbage will begin to appear in the
16th digit.
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