On 15 Sep 2008, at 12:51 pm, Daniel Fischer wrote:
Am Montag, 15. September 2008 02:24 schrieb Cetin Sert:
Hi why do I get?
Buffering. For compiled programmes, stdin and stdout are line-
buffered by
default, so the output doesn't appear until the program finishes.
Either put
hSetBuffering
On 2008 Sep 14, at 20:47, Brandon S. Allbery KF8NH wrote:
On 2008 Sep 14, at 20:24, Cetin Sert wrote:
Hi why do I get?
[EMAIL PROTECTED]:~/lab/exp/1> ./eq
23
23
3
a = b = c = n1-0.8457820374040622n2-0.1542179625959377
As is typical for Unix, filehandles including standard input and
standard
Am Montag, 15. September 2008 02:24 schrieb Cetin Sert:
> Hi why do I get?
Buffering. For compiled programmes, stdin and stdout are line-buffered by
default, so the output doesn't appear until the program finishes.
Either put
hSetBuffering stdout NoBuffering
at the top of main or, better IMO,
ins
Hello,
If we look at these two examples, it appears that the results are reversed:
Prelude> let n o = (-1 `o` 1) in n (-)
0
Prelude> let n o = (-1 `o` 1) in n (+)
-2
Prelude>
we expect (-1 - 1) = -2 and (-1 + 1) = 0, but we get the opposite.
Due to operator precedence, the equations are being
On 2008 Sep 14, at 20:24, Cetin Sert wrote:
Hi why do I get?
[EMAIL PROTECTED]:~/lab/exp/1> ./eq
23
23
3
a = b = c = n1-0.8457820374040622n2-0.1542179625959377
As is typical for Unix, filehandles including standard input and
standard output are line buffered. See hSetBuffering ( http://www.
Hi why do I get?
[EMAIL PROTECTED]:~/lab/exp/1> ./eq
23
23
3
a = b = c = n1-0.8457820374040622n2-0.1542179625959377
when I run
import System.IO
main :: IO ()
main = do
a ← ask "a"
b ← ask "b"
c ← ask "c"
eval a b c
ask v = do
putStr (v ++ " = ")
readLn
eval a b c = do
case delta
