在python中使用两种不同类型的参数调用函数

Question

I am new to Python. I came across a weird case and am not able to figure out what the issue is. I have 2 versions of a function written in Python :-

v1 -

def fileLookUp(fixedPath, version):
    if version:
        targetFile="c:\\null\\patchedFile.txt"
    else:
        targetFile="c:\\null\\vulFile.txt"

#some more code follows

and v2 -

def fileLookUp(fixedPath, version):
    if version:
        print "ok"
    else:
        print "not ok"

#some more code follows

where the parameter fixedPath is a string that is entered and the parameter version is supposed to be an integer value. The 1st function (v1) does not work as expected, while tje second works perfectly. Both the times the function is called as fileLookUp("c:\\\\dir\\\\dir\\\\", 1) .

In the 1st case the error received is :-

fileLookUp("D:\\Celine\\assetSERV\\", 1)
Exception: fileLookUp() takes exactly 2 arguments (1 given)

Please let me know why is the 1st function throwing the exception?

Here is the actual code....

from System.IO import *;

def fileLookUp(fixedPath, version):
if version:
    targetFile="c:\\null\\patchedFile.txt";
else:
    targetFile="c:\\null\\vulFile.txt";
vulFileHandle=open(targetFile,"a+");
temp=fixedPath;
if not Directory.GetDirectories(fixedPath):
    files=Directory.GetFiles(fixedPath);
    for eachFile in files:
        print eachFile;
        hash = Tools.MD5(eachFile);
        print hash;
        vulFileHandle.write(eachFile+'\t'+hash+'\n');
else:
    directory=Directory.GetDirectories(fixedPath);
    for folder in directory:
        if vulFileHandle.closed:
            vulFileHandle=open(targetFile,"a+");
        fixedPath="";
        fixedPath+=folder;
        fixedPath+="\\";
        vulFileHandle.close();
        fileLookUp(fixedPath);
    filess=Directory.GetFiles(temp);
    for eachFilee in filess:
        if vulFileHandle.closed:
            vulFileHandle=open(targetFile,"a+");
        print eachFilee;
        hashh = Tools.MD5(eachFilee);
        print hashh;
        vulFileHandle.write(eachFilee+'\t'+hashh+'\n');
if not vulFileHandle.closed:
    vulFileHandle.close();

it is simply a recursive code to print out the hash of all files in a directory.

Answer 1

You have a call "fileLookUp(fixedPath);" around line 26 or so (just counted roughly) with only one argument sent in. Your definition doesn't allow that. Send in the version in this call, or give a default value to version in the definition.

Answer 2

The way these functions are written, both of them must be called with two arguments. The error message you're getting indicates that one of them is being called with only one argument. In Python, if you want to make an argument optional, you have to explicitly state what value the optional argument should have if it is not provided. For example, since version is to be an int , and you test it with if version: , a good default value could be 0. You could also use None .

def fileLookUp(fixedPath, version=0):
    # etc.

def fileLookUp(fixedPath, version=None):
    # etc.

If 0 is a valid version number, and you want to test for whether a value was actually passed, use the second and test against None specifically:

def fileLookUp(fixedPath, version=None):
    if version is None:
        # etc.

Answer 3

在（格式很差的）完整fileLookUp ，您要递归调用fileLookUp 而不传递version参数。