Challenge: Algorithms having only one line of code.

[Antoni Gual]

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?

[Moneo]

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?
*****

[Antoni Gual]

This way works with 128:

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.

[na_th_an]

I use a replacement for this kind of instruction:

Code:

IF <condition> THEN
   a = <value1>
ELSE
   a = <value2>
END IF

The replacement goes as follows:

Code:

a = (<value1> AND <condition>) + (<value2> AND (NOT <condition>))

Workis well with integer data.

[oracle]

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...

[BinarySHOCK]

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

[whitetiger0990]

this rounds abd is 1 line of code minus the lines that define the variables.

Code:

dec = 1      'how many decimal places to round
num = 2.56   'number to round
PRINT INT(10 ^ dec * num + .5) / 10 ^ dec

This is actually ALauzier's but I just posted it

[Moneo]

An excellent example of how a pure logic statement can replace an IF statement.
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[Moneo]

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:

Code:

IF NUMBER < 0 THEN RFACTOR=-.5 ELSE RFACTOR=.5
RESULT=INT(NUMBER+RFACTOR)

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[whitetiger0990]

Code:

CLS
dec = 2
num = -2.7234
IF NUMBER < 0 THEN rfactor = -.5 ELSE rfactor = .5
PRINT INT(10 ^ dec * num + rfactor) / 10 ^ dec

there