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9 1 1 1. Invalid syntax simply means that the code you have written cannot be interpreted as valid instructions for python. "Syntax" refers to the rules and structures of a language, normally spoken, but also in programming. – ApproachingDarknessFish.
Double check the book example and make sure you have the exact same (tabs, new lines, and all) syntax written. It may look closer to this: def f(x): return x + 2. note the new line and tab. To call this function, on a separate line say: f(5) or replace 5 with whatever parameter you want.
9. Python does not allow empty blocks, unlike many other languages (since it doesn't use braces to indicate a block). The pass keyword must be used any time you want to have an empty block (including in if/else statements and methods). For example, if 3 > 0: print('3 greater then 0') else: pass. Or an empty method:
When i try use print str print str(i) + '/' + str(len(pass_list)) SyntaxError: Invalid syntax
The syntax is changed in new 3.x releases rather than old 2.x releases: for example in python 2.x you can write: print "Hi new world" but in the new 3.x release you need to use the new syntax and write it like this: print("Hi new world")
While it makes sense here to have answers that show problems caused by other kinds of valid syntax, this is an example where the version of Python used causes the syntax not to be valid. This question, fundamentally, is about code becoming invalid because of other parts of the code .
But \A is not valid in a Python string literal! This is invalid: my_regex = "\Afoo". Instead you should do this: my_regex = r"\Afoo". Docstrings are another one to remember: docstrings are string literals too, and invalid \ sequences are invalid in docstrings too! Use r"""raw strings""" for docstrings if they must contain \.
Python expects an indented block after your elif statement, same as for an if, while or else statement. In your example, everything after elif should be indented. answered Feb 21, 2013 at 0:24. ali_m. 74k 28 227 307. Everything is lined up, infact the if statement above it which is if == "1" is the exact copy of this one, yet it still gives an ...
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1. I would recommend logically separating out what you're trying to do. This will make for cleaner code, and, more often than not, code that behaves like you actually want it to. If you want to increment x before printing it, do: x = 1. x += 1. print(x)