Solved

# truncate

Posted on 1998-10-18
1,168 Views
VC5

Is there function that i can truncate
a double of value 3.333333333333333......
to a fewer number of decimal point?

double D = 3.33333333..... // return by some function
double MyNewD = truncate_D_to_2_decimal_point(D) // ???

0
Question by:leowlf
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LVL 3

Expert Comment

ID: 1175414
one way is to do:

double truncate (double D,int n)
{
long multiplied;
double to_return;

multiplied = (D*pow(10,n));         // save only the first n digits after the decimal point . (as a long integer).
to_return  = (multiplied / (pow(10,n));  // return to the trtunced (original ) number.

return (to_return);
}

for example:
in truncate (0.123456789,5) you'll get:
multiplied = 12345
to_return = 0.12345

0

LVL 22

Expert Comment

ID: 1175415
That would work, but you have to understand that because of the limitations of the floating point format, not all decimal numbers will be expressible, thus when you truncate the number, you may see the last digit or two change in value.  This is not problem with arnond's method, this is a limitation of the floating point system.  For this reason, most programmer's don't truncate the value at all.  They may limit the number of digits that print, but will try to retain full precision internally.
0

LVL 3

Expert Comment

ID: 1175416
nietod is right, you can limit the number of digits you print using the format specifications of your favorite output function. for example: in printf() you should use "%.pf".
the 'p' stands for precision, i.e.: how many decimal places to display.

0

LVL 3

Expert Comment

ID: 1175417
leowlf,
did my suggestion help ?
would you like me to post an answer for this question ?
0

Author Comment

ID: 1175418
arnond,
Your suggestion is good enough for me,
Thank you.
0

LVL 3

Accepted Solution

arnond earned 50 total points
ID: 1175419
Thanks.
0

LVL 1

Expert Comment

ID: 24084984
Hi Arnod,
I am using C language because I can not use C++ due to some desing issue.
how do we proceed if the double has this value
UINT s = 10;
int n = 16;
double dx = (double)(s/100.0);
here is dx = 0.10000000000000001
The above logic does not work.
0

LVL 84

Expert Comment

ID: 24085285
> you have to understand that because of the limitations of the floating point format, not all decimal numbers will be expressible, thus when you truncate the number, you may see the last digit or two change in value.
0

LVL 1

Expert Comment

ID: 24104023
Thanks OZO. This is also my understanding.
One concern we are use C code in such a huge platform like Win32 (using VS 2008 compiler). Can this irreguarity be fix by the vendor in near future?
With this I stop with satisfaction.
0

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