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Hi! I'm using LONG DOUBLE data types & I get scientific notation starting at a million.
Is there any way to "force" the program to either:
a) DISPLAY it as a regular number
b) CHANGE it to a regular number
Also, I would imagine that there is a loss in precision with scientific notation. Is there a way to calculate numbers on that scale (maybe up to 1E+36) without any loss in precision using C?
Thanks!
mw
Well you can tweak the width and precision to get more significant digits out
> Is there a way to calculate numbers on that scale (maybe up to 1E+36) without any loss in precision using C?Code:#include <stdio.h>
int main ( void ) {
double d = 12345678901.2345;
printf( "%15.30f\n", d );
return 0;
}
In my math.h, I have
#define LDBL_DIG 18
Basically, long doubles can store 18 decimal digits accurately.
Use a math library like this if you want more digits http://www.swox.com/gmp/
Uh, I don't understand that "%15.30f" in your printf statement. Can you clarify?
Also, is there a way to display it without a decimal?
Thanks!
mw
It's just a extra info how the number is formated.Quote:
Originally Posted by Lionmane
that means: put up to 15 digits on the left side of the point, and up to 30 digit on the right side. something like
000000000000000.000000000000000000000000000000
trailing zeros are disposed.
If I change the printf statement to this:
printf ("%15f\n", d);
It still prints 6 digits on the right side of the decimal. Is there a way to print any number of digits to the left side of the decimal & none on the right?
Thanks!
mw
Not exactly. It means, "use at least 15 characters to represent the whole shebang, but use 30 digits after the decimal point".Quote:
Originally Posted by xErath
Quote:
Originally Posted by Lionmane
[edit]Code:#include <stdio.h>
int main(void)
{
double d = 12345678901.2345;
printf("d = %45.30f\n", d);
printf("d = %.0f\n", d);
return 0;
}
/* my output
d = 12345678901.23450088000000000000000000000
d = 12345678901
*/
Code:#include <stdio.h>
int main(void)
{
double d = 12345678901.2345;
int any_number = 40;
printf("d = '%45.30f'\n", d);
printf("d = '%*.0f'\n", any_number, d);
return 0;
}
/* my output
d = ' 12345678901.23450088000000000000000000000'
d = ' 12345678901'
*/
I hate to beat a dead horse here, but when I use this:
printf ("%45.0f\n", d);
It prints 45 spaces to the left of the decimal whether there are 45 digits or not. Is there a way to truncate the extra spaces?
mw
Reread the second output of the first code fragment of my previous reply.
Heck yeah! Thanks, man!
mw
Yes :) whatever... tricky englishQuote:
Originally Posted by Dave_Sinkula