我知道我只是想念一些愚蠢的东西,但我很沮丧,我只是看不到它。当我对下面的代码运行3个字段的文件“field1; field2; field3”时,这些任务在retrieveTasks方法中正确输出。当该方法返回到主输出已损坏。我知道这很愚蠢,我只是想念它。我已经尝试将CourseName,TaskDescription等改为数组而不是char *。我曾尝试传递fileString作为char *,我已经尝试过使用strcpy而不使用。任何人都可以指出我的方向吗?方法返回后内存被破坏
const int MAX_STRING_LENGTH = 101;
const int MAX_TASK_ITEMS = 255;
const char NEWLINE = '\n';
const char DELIMETER[] = ";";
const char FILENAME[] = "tasks.txt";
struct Task {
char* CourseName;
char* TaskDescription;
char* DueDate;
char FileString[300];
void printTask() {
cout << CourseName << DELIMETER << TaskDescription << DELIMETER << DueDate << endl;
}
void initializeFromFileString(char fileString[]) {
strcpy(FileString, fileString);
CourseName = strtok(FileString, DELIMETER);
TaskDescription = strtok(NULL, DELIMETER);
DueDate = strtok(NULL, DELIMETER);
}
};
struct TaskList {
Task Tasks[MAX_TASK_ITEMS];
int TaskCount;
void initialize() {
TaskCount = 0;
return;
}
void addTask(Task task) {
Tasks[TaskCount] = task;
TaskCount++;
return;
}
void printTasks() {
for(int TaskNum = 0; TaskNum < TaskCount; TaskNum++) {
cout << TaskNum + 1 << ". ";
Tasks[TaskNum].printTask();
cout << endl;
}
return;
}
// Load data from the file. Will return -1 if it fails for any reason.
// Otherwise it returns the number of records read.
int retrieveTasks(const char* fileName) {
int isSuccessfulOpen = 0;
int recordsRead = 0;
ifstream inFile;
isSuccessfulOpen = openFile(inFile, fileName);
if(!isSuccessfulOpen) {
return -1;
}
// Read input file and store in appropriate arrays
while(inFile.eof() == false) {
char fileLine[MAX_STRING_LENGTH * 3];
Task task;
inFile.getline(fileLine, MAX_STRING_LENGTH * 3, NEWLINE);
inFile.ignore(UINT_MAX, NEWLINE);
task.initializeFromFileString(fileLine);
addTask(task);
recordsRead++;
}
inFile.close();
return recordsRead;
}
};
int main() {
bool isFinished = false;
TaskList taskList;
taskList.initialize();
taskList.retrieveTasks(FILENAME);
taskList.printTasks();
return 0;
}
int openFile(ifstream& inFile, const char* fileName) {
inFile.open(fileName);
// Veryify that the file is valid. If not print error message.
if(inFile.is_open() == false) {
cout << "File " << fileName << " does not exist. Please provide a valid file path." << endl;
return 0;
}
return 1;
}
// Open file for writing
int openFile(ofstream& outFile, const char* fileName) {
outFile.open(fileName);
// Veryify that the file is valid. If not print error message and exit.
if(outFile.is_open() == false)
{
cout << "File " << fileName << " does not exist. Please provide a valid file path." << endl;
return 0;
}
return 1;
}
为什么你使用C字符串和库? – 2013-04-28 06:20:37
确实。使用C++技术可以帮助您避免出现三条/五条规则的问题,更不用说使代码更清晰,更易于遵循。 – chris 2013-04-28 06:23:34
我同意。你的本地缓冲区使得copy-ctor和赋值运算符相当平凡,但老实说,这应该使用'std :: string'对象和标准集合类(比如'std :: vector <>')来完成你的任务列表。它会使代码更清洁。 – WhozCraig 2013-04-28 06:31:50