On 20 Sep, 14:35, candide <cand...@free.invalid> wrote: > Let's code a function allowing access to the multiples of a given > integer (say m) in the range from a to b where a and b are two given > integers. For instance, with data input > > a,b,m=17, 42, 5 > > the function allows access to : > > 20 25 30 35 40 > > Each of the following two functions mult1() and mult2() solves the > question : > > # ----------------------------------------- > def mult1(a,b,m): > return (x for x in range(a,b)[(m-a%m)%m:b:m]) > > def mult2(a,b,m): > return range(a,b)[(m-a%m)%m:b:m] > # ----------------------------------------- > > mult2() function returns a list and obviously mult2() needs Python to > allocate memory for this list. What I was wondering is if the same might > be said about mult1(). More precisely, does Python allocate memory for > the whole target list range(a,b)[m-a%m:b:m]?
Yes - range always creates a list (in the 2.x series) -- mult1 creates a list, then returns a generator, so list(mult1(a,b,m)) is identical to mult2(a,b,m). Jon. -- http://mail.python.org/mailman/listinfo/python-list
