strtok / strtok_r困境 | parseNameString

externC char * strtok_r（char * s1，const char * s2，char **持续）; #define MAX_DATA_LEN 128 #define EMPLOYEE_DELIMITER_STR"：" ; #define ATTR_DELIMITER_STR"，" #define ATTR_DELIMITER_CHAR''，'' struct EmployeeInfo { char _name [MAX_DATA_LEN]; size_t _attributeCount; size_t * _attribute; EmployeeInfo（const char * str）：_ attributeCount（0），_ attribute（NULL） { if（！strstr（str，ATTR_DELIMITER_STR）） strcpy（_name，str）; else { char * temp1 = new char [strlen（str）+ 1] ; strcpy（temp1，str）; char * temp2 = temp1; char * holdingBuf [1]; strcpy（_name，strtok_r（temp1，（const char *）ATTR_DELIMITER_STR， holdingBuf））; temp1 + =（ strlen（temp1）+ 1）; _attributeCount = numChars（temp1，ATTR_DELIMITER_CHAR）+ 1; _a ttribute = new size_t [_attributeCount]; memset（_attribute，''\'''，_ attributeCount * sizeof（* _ attribute））; char * stringWithTokens = temp1; char * token; int i; for（i = 0;（token = strtok_r（stringWithTokens，（ const char *） ATTR_DELIMITER_STR，holdingBuf））！= NULL; i ++） { stringWithTokens = NULL; _attribute [i] = atoi（令牌）; } _attributeCount = i; 删除[] temp2; } } virtual~EmployeeInfo（） { if（_attributeCount）delete [] _attribute; } void toString（） { cout<< " \ nEmployee Name：" << _name<< endl; for（size_t i = 0; i< _attributeCount; i ++） cout<< 属性 << i<< ： << _attribute [i]<<结束; } 受保护： static size_t numChars（const char * str，const char delimiter） { size_t retVal = 0; char ch; for（size_t i = 0; ch = str [i] ; i ++） { if（ch == delimiter） retVal ++; } 返回retVal; } }; void parseNameString（const char * nameString） { char * temp1 = new char [strlen（nameString）+ 1]; strcpy（temp1，nameString）; char * temp2 = temp1; char * holdingBuf [1]; list< EmployeeInfo *> empInfoList; char * stringWithTokens = temp1; char * token; while（（token = strtok_r（stringWithTokens，（ const char *） EMPLOYEE_DELIMITER_STR，holdingBuf））！= NULL） { stringWithTokens = NULL; EmployeeInfo * newInfo = new EmployeeInfo（token）; empInfoList.push_back（newInfo）; } delete [] temp2 ; list< EmployeeInfo *> :: iterator itor，endOfList = empInfoList.end（）; for（itor = empInfoList。 begin（）; itor！= endOfList; itor ++） { EmployeeInfo * thisInfo = * itor; thisInfo-> toString（） ; } for（itor = empInfoList.begin（）; itor！= endOfList; itor ++） { EmployeeInfo * thisInfo = * itor; 删除thisInfo; } } int main（int argc，char * argv []） { if（argc< 2） { cerr<< 用法： << argv [0]<< " < inputfile中取代。退出.... << endl; 退出（-1）; } ifstream inFile（argv [1]）; if（！inFile） { cerr<< 读取文件时出错 << argv [1]<< "退出.... << endl; 退出（-2）; } char inputStr [MAX_DATA_LEN + 1]; / * while（inFile>> inputStr） { cout<< **解析字符串 << inputStr<< " ** \ n \ n" ;; parseNameString（inputStr）; cout<< " \ n \ n \ n"; } * / while（inFile.getline（ inputStr，MAX_DATA_LEN）） { cout<< **解析字符串 << inputStr<< " ** \ n \ n" ;; parseNameString（inputStr）; cout<< \ n \ n \ nn; } 退出（0）; } 将整个源列表发布到comp.lang.c新闻组是一个无意中的错误，深感遗憾。 kb***@kaxy.com 写道：我用strtok_r编写了一些代码，用于解析具有两个level的字符串。分隔符。代码工作就像一个魅力，直到有一天突然爆发。你看，我在输入字符串本身上应用strtok，只要变量字符串被传递给函数，一切都是hunky dory。然后有一天有人决定将一个恒定的字符串传递给我的代码----所有的东西都像一张多米诺骨牌一样崩溃了。我相信这是一个陷阱许多程序员可能会失败。我现在已经改变了我的函数的签名，将输入的字符串视为一个常量字符串，我现在制作一个字符串的本地副本并进行操作那。当然，我现在必须确保我释放动态分配的内存。 Ciao KB [我已将我的源列表发布到comp.sources .d] char * strtok（char *，const char *）; strtok插入''\\ \\ 0''代替每个分隔符连续调用。传递一个字符串文字肯定会导致麻烦。传递字符串缓冲区。 ;）事实上，臭名昭着的臭名昭着小丑的书也做了你所做的事！远离他的书，除非你想要练习进行调试。 ;）问候， Jonathan。 - "女性应该附带文件。 - 戴夫 I had written some code using strtok_r for parsing strings with two"levels" of delimiters. The code was working like a charm, until itsuddenly broke one day. You see, I was applying strtok on the inputstring itself, and as long as a variable string was passed to thefunction, everything was hunky dory. Then one day somebody decided topass a constant string to my code ---- and everything came collapsingdown like a domino.I believe that this is a pitfall into which many programmers may fall.I have now changed the signature of my function to treat the inputstring as a constant string, and I now making a local copy of thestring and operating on that. Of course, I now have to ensure that Ifree up the dynamically allocated memory.CiaoKB[I have posted my source listing to comp.sources.d] 解决方案 Modified source listing for my program using strtok_r in which I make alocal copy of the input string and operate strtok on that// Program that parses strings of the form:// Name, SS#, Emp#, Other#::Name, SS#, Emp#, Other#::Name, SS#, Emp#,Other#// and extracts the name and other attributes associated with anemployee#include <stdio.h>#include <iostream.h>#include <fstream.h>#include <list.h>extern "C" char *strtok_r(char *s1, const char *s2, char **lasts);#define MAX_DATA_LEN 128#define EMPLOYEE_DELIMITER_STR ":"#define ATTR_DELIMITER_STR ","#define ATTR_DELIMITER_CHAR '',''struct EmployeeInfo{char _name[MAX_DATA_LEN];size_t _attributeCount;size_t *_attribute;EmployeeInfo(const char *str):_attributeCount(0), _attribute(NULL){if(!strstr(str,ATTR_DELIMITER_STR))strcpy(_name, str);else{char *temp1 = new char[strlen(str) + 1];strcpy(temp1, str);char *temp2 = temp1;char* holdingBuf[1];strcpy(_name, strtok_r(temp1, (const char *) ATTR_DELIMITER_STR,holdingBuf));temp1 += (strlen(temp1) + 1);_attributeCount = numChars(temp1, ATTR_DELIMITER_CHAR) + 1;_attribute = new size_t[_attributeCount];memset(_attribute, ''\0'', _attributeCount*sizeof(*_attribute));char* stringWithTokens = temp1;char* token;int i;for(i =0;(token = strtok_r(stringWithTokens, (const char *)ATTR_DELIMITER_STR, holdingBuf)) != NULL;i++){stringWithTokens = NULL;_attribute[i] = atoi(token);}_attributeCount = i;delete [] temp2;}}virtual ~EmployeeInfo(){if(_attributeCount) delete [] _attribute;}void toString(){cout << "\nEmployee Name: " << _name << endl;for(size_t i = 0; i < _attributeCount; i++)cout << "Attribute " << i << ": " << _attribute[i] << endl;}protected:static size_t numChars(const char* str, const char delimiter){size_t retVal = 0;char ch;for(size_t i = 0; ch = str[i]; i++){if(ch == delimiter)retVal++;}return retVal;}};voidparseNameString(const char* nameString){char *temp1 = new char[strlen(nameString) + 1];strcpy(temp1, nameString);char *temp2 = temp1;char* holdingBuf[1];list<EmployeeInfo*> empInfoList;char* stringWithTokens = temp1;char* token;while((token = strtok_r(stringWithTokens, (const char *)EMPLOYEE_DELIMITER_STR, holdingBuf)) != NULL){stringWithTokens = NULL;EmployeeInfo *newInfo = new EmployeeInfo(token);empInfoList.push_back(newInfo);}delete [] temp2;list<EmployeeInfo*>::iterator itor, endOfList=empInfoList.end();for(itor = empInfoList.begin(); itor != endOfList; itor++){EmployeeInfo* thisInfo = *itor;thisInfo->toString();}for(itor = empInfoList.begin(); itor != endOfList; itor++){EmployeeInfo* thisInfo = *itor;delete thisInfo;}}intmain(int argc, char* argv[]){if(argc < 2){cerr << "usage: " << argv[0] << " <inputfile>. Exiting ...." <<endl;exit(-1);}ifstream inFile(argv[1]);if(!inFile){cerr << "Error reading file " << argv[1] << " Exiting ...." <<endl;exit(-2);}char inputStr[MAX_DATA_LEN + 1];/*while(inFile >> inputStr){cout << "** Parsing string " << inputStr << " **\n\n";parseNameString(inputStr);cout << "\n\n\n";}*/while(inFile.getline(inputStr, MAX_DATA_LEN)){cout << "** Parsing string " << inputStr << " **\n\n";parseNameString(inputStr);cout << "\n\n\n";}exit(0);} The posting of the entire source listing to the comp.lang.c newsgroupwas an inadvertent mistake which is deeply regretted. kb***@kaxy.com wrote: I had written some code using strtok_r for parsing strings with two "levels" of delimiters. The code was working like a charm, until it suddenly broke one day. You see, I was applying strtok on the input string itself, and as long as a variable string was passed to the function, everything was hunky dory. Then one day somebody decided to pass a constant string to my code ---- and everything came collapsing down like a domino. I believe that this is a pitfall into which many programmers may fall. I have now changed the signature of my function to treat the input string as a constant string, and I now making a local copy of the string and operating on that. Of course, I now have to ensure that I free up the dynamically allocated memory. Ciao KB [I have posted my source listing to comp.sources.d]char* strtok (char*, const char*);strtok works by inserting ''\0'' in place of the delimiter eachsuccessive call. Passing a string literal will definitelycause trouble. Pass a string buffer. ;)In fact, Herb-the-infamous-clown''s book does what you did too!Stay away from his books, unless you want an exercisein debugging. ;)Regards,Jonathan.--"Women should come with documentation." - Dave 这篇关于strtok / strtok_r困境的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！上岸，阿里云！