(no) fast boolean evaluation ?
Paddy
paddy3118 at googlemail.com
Sat Aug 4 17:18:28 CEST 2007
On Aug 2, 10:47 pm, Stef Mientki <S.Mientki-nos... at mailbox.kun.nl>
wrote:
> hello,
>
> I discovered that boolean evaluation in Python is done "fast"
> (as soon as the condition is ok, the rest of the expression is ignored).
>
> Is this standard behavior or is there a compiler switch to turn it on/off ?
>
> thanks,
> Stef Mientki
The following program shows a(clumsy)? way to defeat the short-
circuiting:
def f(x):
print "f(%s)=%s" % ('x',x),
return x
def g(x):
print "g(%s)=%s" % ('x',x),
return x
print "\nShort circuit"
for i in (True, False):
for j in (True, False):
print i,j,":", f(i) and g(j)
print "\nShort circuit defeated"
for i in (True, False):
for j in (True, False):
print i,j,":", g(j) if f(i) else (g(j) and False)
The output is:
Short circuit
True True : f(x)=True g(x)=True True
True False : f(x)=True g(x)=False False
False True : f(x)=False False
False False : f(x)=False False
Short circuit defeated
True True : f(x)=True g(x)=True True
True False : f(x)=True g(x)=False False
False True : f(x)=False g(x)=True False
False False : f(x)=False g(x)=False False
- Paddy.
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