07-02-2003, 02:53 AM
Adding .5 before INT does round to the next integer, the same as CINT does. INT is a floor function. Maybe your original code is in another language?
07-02-2003, 02:53 AM
07-02-2003, 03:03 AM
Yes, the original code was in PICKBasic, which is identical for this one line of code.
What do you mean by a "floor function"?
Have you been able to determine why the formula fails for N=128 without the rounding?
*****
What do you mean by a "floor function"?
Have you been able to determine why the formula fails for N=128 without the rounding?
*****
07-02-2003, 03:15 AM
This way works with 128:
So i imagine it's a QB bug...
A floor function rounds towars the smaller integer.
Code:
DEFLNG A-Z
a! = LOG(n) / LOG(2)
IF 2 ^ a! = n THEN PRINT n; " is a power of 2"So i imagine it's a QB bug...
A floor function rounds towars the smaller integer.
07-02-2003, 03:25 AM
I use a replacement for this kind of instruction:
The replacement goes as follows:
Workis well with integer data.
Code:
IF <condition> THEN
a = <value1>
ELSE
a = <value2>
END IFThe replacement goes as follows:
Code:
a = (<value1> AND <condition>) + (<value2> AND (NOT <condition>))Workis well with integer data.
07-02-2003, 07:28 AM
I've got a function for statlib somewhere that does correct rounding (eg .5 and up rounds up, and <.5 rounds down) because the QB ones wern't working for all decimal values. I'll try to find it...
07-02-2003, 11:10 AM
Here's a crappy rounding thingy
N = 3445 ' Number To Round
Dire = 0 ' Direction to round, 1 = Up, 0 = Down
Rounded = CINT(N + ((Dire = 0) - (Dire = 1)) * .5)
PRINT Rounded
N = 3445 ' Number To Round
Dire = 0 ' Direction to round, 1 = Up, 0 = Down
Rounded = CINT(N + ((Dire = 0) - (Dire = 1)) * .5)
PRINT Rounded
07-02-2003, 06:29 PM
this rounds abd is 1 line of code minus the lines that define the variables.
This is actually ALauzier's but I just posted it
Code:
dec = 1 'how many decimal places to round
num = 2.56 'number to round
PRINT INT(10 ^ dec * num + .5) / 10 ^ dec07-02-2003, 08:29 PM
An excellent example of how a pure logic statement can replace an IF statement.
*****
*****
07-02-2003, 08:47 PM
Let's not forget that positive numbers are rounded UP using .5, and negative numbers are rounded DOWN using -.5.
So, the simplest way is to test the sign of the number to be rounded, and set the rounding factor to .5 or -.5 accordingly. Something like this:*****
So, the simplest way is to test the sign of the number to be rounded, and set the rounding factor to .5 or -.5 accordingly. Something like this:
Code:
IF NUMBER < 0 THEN RFACTOR=-.5 ELSE RFACTOR=.5
RESULT=INT(NUMBER+RFACTOR)07-02-2003, 08:58 PM
Code:
CLS
dec = 2
num = -2.7234
IF NUMBER < 0 THEN rfactor = -.5 ELSE rfactor = .5
PRINT INT(10 ^ dec * num + rfactor) / 10 ^ dec