C++ 损坏的双链表

Question

我是来自 Java 的 c++ 新手，我正面临 c++ I/O 的问题。我做了一个小程序来说明我的问题。主要思想是拥有 3 维数组并通过读取具有特定文本格式的文本文件来初始化它的值。该文件包含第一个维度的级别数 (L) 以及第二个和第三个维度 (array[L][N][N]) 的行和列的大小 (N)。

编辑：为了执行我的程序，我使用g++ -std=c++0x program.cpp

这是程序：

#include <iostream>
#include <string>
#include <fstream>

using namespace std;

int num_of_stages;
int N ;
string ***stages;

void readFile(string file_name){
    string line;
    int pos;

    ifstream file (file_name);

    if (file.is_open()){

        getline (file,line);                    // L=XXX
        pos = line.find("=");
        line = line.substr(pos+1);
        num_of_stages = atoi(line.c_str());
        cout << "L=" << num_of_stages << endl;

        getline (file,line);                    // N=XXX
        pos = line.find("=");
        line = line.substr(pos+1);
        N = atoi(line.c_str());
        cout << "N=" << N << endl;      

        // setting the stages array with dimensions stages[L][N][N]
        stages = new string**[num_of_stages];
        for(int i = 0 ; i < N ; i ++){
            stages[i] = new string*[N];
            for(int j = 0 ; j < N ; j++){
                stages[i][j] = new string[N];
            }
        }

        for(int i = 0 ; i < num_of_stages ; i++){
            getline (file,line);                // Level comment
            cout << "Level " << line << endl; 
            for(int j = 0 ; j < N ; j++){
                for(int k = 0 ; k < N ; k++){
                    string currentValue ;
                    file >> currentValue;
                    stages[i][j][k] = currentValue;
                    cout << currentValue << " " ;
                }
                cout << endl;
            }
            cout << endl << endl;
            // I am guessing that this getLine() 
            // is necessary to read the end of the line
            // if I try remove it everything goes wrong
            // everytime I print the line's values is ' ' (empty)
            getline (file,line);
            // remove any string value to detect end of file        
            // so the if statement below will be true
            line.clear();
        }
        getline (file,line);                    // END OF MAZE comment
        cout << line << endl;
        if(line != "END OF MAZE"){
            cout << "Wrong file format" << endl;
            file.close();
            exit(0);
        }
        file.close();
    }else{
        cout << "Missing file with name : " << file_name << endl;
        exit(0);
    }
    // program never reach that point
    cout << "Bla bla bla" << endl;
}

int main(){
    readFile("file1.maz");
    return 0;
}

我的文件有这种文本格式：

L=5
N=16
LEVEL 1
T2 T3 T1 T2 E E T1 E E T2 T2 T1 T2 T1 T2 T1 
T1 T1 T2 T1 T2 T3 T2 T3 T2 T2 T3 T3 T1 E T3 T1 
T2 E T1 T1 T3 T2 T1 T3 E T3 T3 E T1 E T1 T1 
T1 T3 T2 E T1 T1 T2 E T2 T2 T2 T1 T3 T3 T1 T3 
E E T1 T3 T1 T3 T1 T2 T2 T1 T1 E T2 T1 T1 T3 
T2 T1 T1 T3 T3 E T2 T3 T3 T2 T1 E T3 T2 T1 T2 
T2 T1 T3 T2 T2 E T2 E E E T3 T3 T2 T3 T3 T1 
E E E T2 E E T3 E T1 T2 T1 T2 T1 T3 T3 E 
T3 T2 T1 T3 E T1 E T2 T3 T1 T2 T2 T3 T2 E T1 
T2 T3 E E T2 T2 E T3 E E T3 T1 T3 T1 E T3 
T1 T1 E T1 T1 T3 T3 T1 E T2 T3 T1 T3 T3 T3 E 
T3 T1 T1 T1 T2 T2 T3 T1 T3 T2 E T2 T1 E T1 E 
T3 E T2 E T1 T1 T3 T2 T3 T2 T3 E T3 T3 T2 T2 
T1 T1 E T3 T1 E T2 T1 T3 T1 T1 T2 T2 T2 T1 T2 
T3 E T3 T3 T3 T2 T1 T1 E E E T2 T1 T2 E T3 
T1 E T3 T2 E E T3 E E E T3 T1 E T2 T2 E 
LEVEL 2
T2 T2 T3 T2 T1 T1 T3 T3 T1 E T1 T2 T3 T2 E T1 
T1 T3 T2 E T2 E T2 T2 T3 T1 T3 T1 E T3 T3 T3 
T3 T3 T3 E T3 T3 E T3 T1 T1 T1 T2 T2 T3 T2 T1 
T2 T3 T1 T2 T1 E E T2 T1 T1 T3 T2 E T1 T2 T3 
T2 T2 T3 E T3 T3 T1 T1 E T3 T3 T2 T3 E T1 E 
T2 T3 T3 T2 E E E T2 T1 T3 E T2 T1 T3 E T1 
T3 T3 E T1 E T3 T1 T1 T3 T3 E T3 E T3 T1 T2 
T2 E T1 E T1 T3 T1 T3 E T3 T1 E T1 T3 T1 E 
T3 E E T1 T3 T2 T2 E T3 T2 T2 T1 E T1 T2 T2 
T3 T3 E E T2 T1 T3 T1 E T3 E E E E T1 E 
T3 E E T2 E E T2 T1 T2 T1 T1 T2 T3 T2 T3 T1 
T3 E T2 T3 T1 T3 T3 T2 E T1 T2 T2 T2 T2 T1 E 
T2 T3 T1 T3 T2 T3 T2 T3 T1 T1 T3 T3 T2 T2 E T3 
T3 T3 T2 T3 T1 T3 T1 T1 T2 E E T3 E T1 T2 T2 
T1 T2 T2 T2 T1 T1 T1 E T1 T2 T2 E T1 T3 T1 T3 
E T2 E E T3 T3 E T1 T1 T2 E T1 T3 E T3 T3 
LEVEL 3
T3 E T1 T3 T3 T3 T2 T3 T1 E T1 T1 T1 T2 E T1 
T1 E E T2 T1 T3 T1 E T1 T3 T3 T3 T3 T2 T1 T1 
T2 T3 T1 T2 T3 E T3 T1 T3 E T1 T3 T1 T2 E T3 
T1 T2 T1 E T2 T1 T1 T3 T2 T2 T2 T2 T3 T3 E T1 
T3 T2 T3 E T3 T1 T1 T2 E T2 E T3 T1 T1 E T1 
T3 E T2 T3 T2 E E T2 T1 E T1 E T2 E T3 T3 
E T2 T1 T3 E T1 E E T3 T1 T1 T3 T2 T1 E E 
T1 T3 T2 T3 T1 T2 T1 E T2 T3 T3 T1 T3 T2 T3 T2 
E T2 T1 T1 E T1 T2 T3 E T2 T1 T3 E T3 E T2 
T3 E T1 T3 T2 E T1 T1 E T2 T3 T3 T3 T2 E E 
T2 E E T2 T3 T3 T1 T2 T1 T2 T1 T3 T2 T2 T2 T2 
T1 E T2 T3 T2 E T2 T2 T1 T2 T1 E T2 E T2 T3 
T2 E T1 E E T1 T3 T1 E T1 E T1 T1 E T2 T2 
T2 T3 T3 T1 E T1 T1 E T1 T3 T3 T3 E T2 T2 E 
T2 T1 T1 T1 T2 E T1 T3 T3 T3 T2 E T3 E T2 E 
E T2 T1 T2 E T1 T3 T1 T3 T1 T3 T2 T1 T1 E T3 
LEVEL 4
E E E T1 T2 T1 T1 E T1 T3 T2 E E T1 T3 T2 
T2 T2 E T3 T3 E E T1 T3 E E T1 T2 T3 T2 E 
T2 T3 T3 T3 T3 E T2 T2 T1 T1 T1 T2 T3 E T2 T1 
T2 T1 E T2 T3 T3 T3 E T1 T3 T2 T2 T2 T1 E T3 
T3 T2 T3 E T2 E T2 E T2 T1 T1 T3 E T2 E T2 
E T1 T1 T1 E E E T2 T1 T1 T3 T1 T2 E T3 T3 
E T3 E T1 E T2 T2 T1 T2 E T3 T3 T3 T1 T3 T3 
E T2 T1 T1 T2 T2 T3 E E T1 T3 T3 T3 T1 T2 E 
T3 T3 T2 E T3 T2 E T1 T2 E E T2 T2 T1 E E 
T3 E E T2 T3 T2 T1 T2 T3 T1 E T1 T3 T2 T1 T2 
E T2 T1 T2 T3 T2 E T3 T3 T1 E T3 T3 T3 T3 T1 
E T2 T1 T1 T1 T3 E T2 E T3 T1 T1 T2 T2 E E 
T3 T3 T3 T2 E T1 T1 T3 T3 T3 E T3 T3 T3 T1 T1 
T2 T1 T1 T2 E E E T3 T1 T3 T3 T1 E E T1 T2 
T1 T2 T2 E T2 T1 T3 T3 T1 T1 T2 T3 T3 T1 T2 T1 
T1 T2 T1 E T1 T3 T2 T1 T1 T2 E E T1 E T3 T1 
LEVEL 5
T1 T1 T3 T1 T2 T3 E T2 T2 T1 T3 T2 T2 E T1 T1 
T3 E T3 E T2 T1 T1 T3 T2 T3 T1 T1 T3 T2 T3 T1 
T2 T3 T1 T2 T2 E T1 T3 E T2 T2 T2 T3 T3 T2 T2 
T1 E T1 T1 T2 T1 T3 E T3 E T1 T1 T3 T3 T1 T3 
T1 T1 T3 T2 T2 E E E E T2 E T1 T1 T2 E T2 
E T2 E E T3 E T1 E E T1 T3 T2 E E T2 E 
T2 T3 T3 E T3 T2 T2 T3 T1 E E E T3 T2 T3 T2 
T1 T1 T1 T1 E T2 E E T1 T3 T2 T1 T3 E E E 
T1 T1 E T1 T3 E E E T1 T1 T1 T3 T1 T2 T3 T3 
T2 T2 T3 T3 E T2 T1 T3 T2 T1 T3 T1 T1 T2 T3 E 
T3 T1 T3 T3 T3 E T1 T1 E T3 T2 E T2 T2 E T1 
T1 T3 E T1 T2 T2 T2 T2 T3 T3 E E T3 T1 T1 T2 
T1 E E T3 T2 T2 T1 T2 T2 T2 T2 T1 T3 T3 T3 E 
T3 E E T2 T1 T3 T1 T2 E E T3 E T3 T2 T1 E 
T1 T3 T1 T3 T2 E E T2 T2 T2 E T1 T2 E T1 T1 
T2 T3 T1 T3 T1 T1 T2 T2 E T1 T2 T1 T2 E E T1 
END OF MAZE

当我使用上面的文件示例执行我的程序时，我收到了这个错误：

Error in `./a.out': corrupted double-linked list: 0x0000000001174280 ***

这是 gdb 回溯：

(gdb) backtrace
#0  0x00007ffff7531cc9 in __GI_raise (sig=sig@entry=6)
    at ../nptl/sysdeps/unix/sysv/linux/raise.c:56
#1  0x00007ffff75350d8 in __GI_abort () at abort.c:89
#2  0x00007ffff756e394 in __libc_message (do_abort=do_abort@entry=1, 
    fmt=fmt@entry=0x7ffff767cb28 "*** Error in `%s': %s: 0x%s ***\n")
    at ../sysdeps/posix/libc_fatal.c:175
#3  0x00007ffff75790f7 in malloc_printerr (action=<optimized out>, 
    str=0x7ffff7678bfc "corrupted double-linked list", ptr=<optimized out>)
    at malloc.c:4996
#4  0x00007ffff757a7a7 in _int_free (av=0x7ffff78b9760 <main_arena>, p=<optimized out>, 
    have_lock=0) at malloc.c:3996
#5  0x00007ffff7b898be in std::basic_filebuf<char, std::char_traits<char> >::_M_destroy_internal_buffer() () from /usr/lib/x86_64-linux-gnu/libstdc++.so.6
#6  0x00007ffff7b89d47 in std::basic_filebuf<char, std::char_traits<char> >::close() ()
   from /usr/lib/x86_64-linux-gnu/libstdc++.so.6
#7  0x00007ffff7b8b3ad in std::basic_ifstream<char, std::char_traits<char> >::close() ()
   from /usr/lib/x86_64-linux-gnu/libstdc++.so.6
#8  0x00000000004018b9 in readFile(std::string) ()
#9  0x0000000000401a3d in main ()

我看过其他关于当前问题的帖子，其中一些是关于摆脱数组索引的，但我已经测试了所有内容，而且我总是在界限内。那么我阅读文件的方式有问题吗？ 我希望这不是一个愚蠢的问题。 提前致谢。

Answer 1

你要越界了：

stages = new string**[num_of_stages];
    for(int i = 0 ; i < N ; i ++) {
        stages[i] = new string*[N];

如果 num_of_stages < N 您正在通过数组访问。

您应该使用std::vector 而不是处理原始指针并自己使用new/delete。它有一个at 函数来进行边界检查。