脚本在解释器中运行奇怪。 -i标志有什么影响？

Question

I recently made a program which prints a word diagonally. Whenever I go to the console/python interpreter and type

python3 "xxx.py" , it will just continue onto the next line and won't do anything.

However, if I do: python3 -i "xxx.py" it enters python and lets me enter inputs for my program

Why is this happening?

My code (copied from comment below):

def diagonal(text, right_to_left = False):
    #Code for diagonal

diagonal()

I got this error:

Traceback (most recent call last): File "question1.py", line 17, in <module> diagonal() 
TypeError: diagonal() missing 1 required positional argument: 'text'

Answer 1

The flag -i tells python to process the script, and then enter interactive mode. Without the -i python will just process the script and then exit.

A script might define functions, classes etc, but not call them. If you want the script to do something, you must have at least one line in your script that calls a function.

The usual pattern is:

#class and function definitions 
def print_diagonal(x):
    #code for diagonal

def main():
    #code for running the program
    word = input()
    print_diagonal(word)

#run the program
main()

The error you get is because the diagonal function has one required argument, the text. You need to supply this argument in some way. You could use an input function in the code (as in my example), or you could use the command line by import sys , and reading sys.argv[] . The python documentation has examples of this