Question

The code is written in python 3.3 and only works for the first if statement, not acknowledging the other elif statements even if the if statement is wrong.

calccircle()

What data do you know? radius

Enter Diameter def calccircle():
      x = input("What data do you know? ")
      if x == "Diameter" or "diameter":
          a = int(input("Enter Diameter "))
          print("Circumference is", a * math.pi)
          print("Area is", math.pi * math.pow(a/2,2))
          print("Radius is:",a/2)
      elif x == "Radius" or "radius":
          b = input("Enter radius: ")
          print("Circumference is", b * 2 * math.pi)
          print("Area is", math.pi * math.pow(b,2))
          print("Diameter is", b * 2)
      elif x == "area" or "Area":
          c = input("Enter area: ")
          print("Circumference is", ((math.sqrt(c))/math.pi) * b * 2 * math.pi)
          print("Diameter is", math.sqrt(c) * math.pi * 2)
          print("Radius is", math.sqrt(c) * math.pi)
      elif x == "circumference" or "Circumference":
          d = input("Enter Circumference: ")
          print("Area is", math.pi * math.pow(d/math.pi,2))
          print("Diameter is", d/math.pi * 2)
          print("Radius is", d/math.pi)

It displays the input(“Enter diameter: “) and doesn’t pay attention to what I write or the if statements.

calccircle()
What data do you know? radius
Enter Diameter

Notice I wrote radius, and the input(“Enter radius: “) is supposed to run, but it doesn’t. Please help.

Answer 1

Your problem is this:

 if x == "Diameter" or "diameter":

looks to python like:

 if (x == "Diameter") or "diameter":

which, when x != "Diameter", is like:

if "diameter":

which will always go through.

Python treats False, None, "", [], {}, ... and the like as False in the context of if statements (or if you call bool on them, or various other places), and most everything else as True. This is often handy, but can combine with a slight confusion about the or statement to make lots of people not used to Python make this mistake.

Instead, do one of these:

if x == "Diameter" or x == "diameter":  # most direct translation
if x in {"Diameter", "diameter"}:  # very slightly faster, a little less typing
if x.lower() == "diameter":  # also allows DIAmeter, etc

---

It’s also worth noting that if you wrote something like

if x == ("Diameter" or "diameter"):

this would be the same as

if x == "Diameter":

since "Diameter" or "diameter" sees that "Diameter" is Trueish, and so returns that.