[Tutor] SyntaxError Message
alan.gauld at btinternet.com
Tue Jul 8 15:40:12 CEST 2014
On 08/07/14 02:41, Pamela Wightley wrote:
> Hi All,
> I have no programming experience and am trying to teaching myself
> python. Am trying to replicate the code below but I get the error
> message below highlighted in yellow:
Unfortunately I can't see anything in yellow, probably due to enmail
losing the formatting.
But the code below has several errors and looks like its part
of a bigger program. It also looks completely unrelated to the
errors you show us further down.
> choice = input("Choose your option: ")
You seem to be using Python 2 in which case using input() is
frowned upon. Its better to use raw_input() and convert the
resultant string to a number using int() or float() as needed.
[ In Python 3 input() has been rwe oved and raw_input renamed to input()...]
> if choice == 1:
indentation(spacing) is very importanbt in Python.
You should only indent the code inside a code block that follows a
structural statement such as if/for/while/def/class etc.
Indenting the if statement after the input() will raise an error.
> add1 = input("Add this: ")
But this should e indented because its part of the if block.
> add2 = input("to this: ")
> print add1, "+", add2, "=", add1 + add2
> elif choice == 2:
> sub2 = input("Subtract this: ")
> sub1 = input("from this: ")
> print sub1, "-", sub2, "=", sub1 - sub2
> elif choice == 3:
> mul1 = input("Multiply this: ")
> mul2 = input("with this: ")
> print mul1, "*", mul2, "=", mul1 * mul2
> elif choice == 4:
> div1 = input("Divide this: ")
> div2 = input("by this: ")
> print div1, "/", div2, "=", div1 / div2
> elif choice == 5:
> loop = 0
> Any assistance appreciated.
> ERROR MESSAGE
>>>> return input (1)
> File "<stdin>", line 1
> SyntaxError: 'return' outside function
This seems completely unrelated to the choice code?
You can only use return inside a function (ie a block
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