Python 只替换不是 C++ 注释的字符串

Question

我只想处理 NOT C++ 注释的字符串，这里是找出 C++ 注释的模式:

pattern = re.compile(r'//.*?$|/\*.*?\*/|\'(?:\\.|[^\\\'])*\'|"(?:\\.|[^\\"])*"', re.DOTALL | re.MULTILINE)

但是，我不知道如何让它按照我的意图工作。

# Python 3.4.2
s = '''
/****
C++ comments
  //pResMgr->CreateDialogEx();
****/
//pResMgr->CreateDialogEx();
/*//pResMgr->CreateDialogEx();*/

// real code, I want to replace only this following line of code
pResMgr->CreateDialogEx();
'''

newS = s.replace('CreateDialogEx', 'Create')
print(newS)

我的预期输出是:

/****
C++ comments
  //pResMgr->CreateDialogEx();
****/
//pResMgr->CreateDialogEx();
/*//pResMgr->CreateDialogEx();*/

// real code, I want to replace only this following line of code
pResMgr->Create();

Answer 1

没有测试它，但它适用于您的情况，并且从根本上应该可以工作。它基本上遍历文本查找换行符，//或/*，然后处理案例。非常简单，没有正则表达式。

source_code = '''//pResMgr//->CreateDialogEx();'''

def indexOf(string, character):
    return string.index(character) if character in string else 9999999

def replaceNotInComments(string, searchFor, replaceWith):
    result = ''
    nextBreak = 0
    while True:
        nextBreak = min(indexOf(string, '\n'),
                        indexOf(string, '/*'),
                        indexOf(string, '//'))
        if nextBreak == 9999999:
            result += string.replace(searchFor, replaceWith);
            break
        result += string[0:nextBreak].replace(searchFor, replaceWith);

        if nextBreak == indexOf(string, '\n'):
            string = string[nextBreak+1:]

        if nextBreak == indexOf(string, '/*'):
            string = string[nextBreak+2:]
            result += '/*'+string[0:indexOf(string, '*/')+2]
            string = string[indexOf(string, '*/')+2:]

        if nextBreak == indexOf(string, '//'):
            string = string[nextBreak+2:]
            if result != '':
                result += '\n'  
            result += string[0:indexOf(string, '\n')+1]
            string = string[indexOf(string, '\n')+1:]


    return result

result = replaceNotInComments(source_code, 'CreateDialogEx', 'Create')
print(result)