如何在python中将文件名作为函数参数传递？

def process_file(filename):
    # 处理文件的代码
    print("处理文件：" + filename)

file_name = "example.txt"
process_file(file_name)

import sys

def process_file(filename):
    # 处理文件的代码
    print("处理文件：" + filename)

file_name = sys.argv[1]
process_file(file_name)

python script.py example.txt

import tkinter as tk
from tkinter import filedialog

def process_file(filename):
    # 处理文件的代码
    print("处理文件：" + filename)

root = tk.Tk()
root.withdraw()

file_name = filedialog.askopenfilename()
process_file(file_name)