> def how_many_times(): > x, y = 0, 1 > c = 0 > while x != y: > c = c + 1 > x, y = roll() > return c, (x, y)
Since I haven't seen it used in answers yet, here's another option using our
new walrus operator
def how_many_times():
roll_count = 1
while (rolls := roll())[0] != rolls[1]:
roll_count += 1
return (roll_count, rolls)
--
https://mail.python.org/mailman/listinfo/python-list
