Count nb call of a function, without global var or decorator

[Méta-MCI]

Hi!

Example, with meta-data (attributs of function) :

def ff(this):
try:
this.count=this.count+1
except:
this.count=1
a=1
b=2
c=a+b

ff(ff)
fa=ff
ff(ff)
fa(fa)
print ff.count

How to improve that?
@-salutations

Michel Claveau

[Méta-MCI]

Re!

I can do :

def ff():
this=ff
try:
this.count=this.count+1
except:
this.count=1
a=1
b=2
c=a+b

ff()
fa=ff
ff()
fa()

print ff.count
But that use, inside the function, the litteral name of the function; and I
want no use litteral name (inside)

@+

Michel Claveau

[Bjoern Schliessmann]

Méta-MCI wrote:

Example, with meta-data (attributs of function) :

Apart from asking what counting "nb call" of a function means, I
wonder why you didn't use an iterator?

@-salutations

@-less Regards,
Björn

--
BOFH excuse #65:

system needs to be rebooted

[Méta-MCI]

Re!

>>why you didn't use an iterator?

If the iterator is extern (to the function), it's like decorator, or global
var.
If it is internal, it's huge, compare to this.count=this.count+1 (or
this.count+=1)
@+

Michel Claveau

[Bjoern Schliessmann]

Méta-MCI wrote:

If the iterator is extern (to the function), it's like decorator,
or global var.

Please excuse me, I don't understand your point. I'm not even sure
if both of us speak of the same iterators.

If it is internal, it's huge, compare to this.count=this.count+1
(or this.count+=1)

In which way huge? If you want top notch performance you should use
a C extension anyway ...

Regards,
Björn

--
BOFH excuse #339:

manager in the cable duct

[Duncan Booth]

"Méta-MCI" <en**************@XmclaveauX.comwrote:

Example, with meta-data (attributs of function) :

def ff(this):
try:
this.count=this.count+1
except:
this.count=1
a=1
b=2
c=a+b

ff(ff)
fa=ff
ff(ff)
fa(fa)
print ff.count

How to improve that?

If I've managed to guess what you are asking, you want to use a class:

>>class Ff:

def __init__(self):
self.count = 0
def __call__(self, notused):
self.count += 1
a, b = 1, 2
c = a+b
return c

>>ff = Ff()
fa = Ff()
ff(ff)

3

>>ff.count

1

>>fa.count

0