If the problem with else else is related to raw_input in Python
I am currently following the Zed Shaw Python book, learning the Python Hard Way and learning functions. I decided to follow some of the extra credit exercises that came with the tutorial and added a post about the current IF ELSE thread. This is the code I have below.
print "How much bottles of water do you have?"
water = raw_input("> ")
print "How many pounds of food do you have?"
food = raw_input("> ")
if food == 1:
def water_and_food(bottles_of_water, food):
print "You have %s bottles of water" % bottles_of_water
print "And %s pound of food" % food
else:
def water_and_food(bottles_of_water, food):
print "You have %s bottles of water" % bottles_of_water
print "And %s pounds of food" % food
water_and_food(water, food)
I want to do this. If the user enters they have 1 pound of food, it will display "You have 1 pound of food." If they enter they have 2 pounds of food or more, the display will show "You have 2 pounds of food", the difference in pounds being single or multiple.
However, if I put 1, it still displays "You have 1 pound of food", however, if I directly assign the number to the water and food variables, it works.
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The return value raw_input
is a string, but when you check the power value you are using int. However, if food == 1
it can never be True
, so the stream is always plural by default.
You have two options:
if int(food) == 1:
The above code will cast food
to integer type, but throws an exception if the user does not enter a number.
if food == '1':
The above code checks for the string '1', not an integer (note the surrounding quotes).
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In Python 2.x, raw_input returns a string. After looking at your code, you can also use input , which returns an integer. I would think this would be the most explicit option using Python2.
Then you can treat food as int throughout your code using% d instead of% s. When you enter a non int, your program throws an exception.
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