Question

我有一个双指针char数组，它将存储字符串行。我的函数是解析从文件读取的行，然后将每个解析的项存储到名为“intermediate.txt”的不同文件中。我希望解析的每个项目都有一个新行添加到其字符串的末尾，因此在中间文件中，每个解析的项目都在它自己的行上。

我正在尝试使用strcat，除了brokenUpLine [4]之外，它似乎与每一行都能正常工作。（我故意在第3个索引处跳过元素）。此时brokenUpLine [4]正在存储“错误代码”。我想在字符串的末尾添加一个换行符，它会一直崩溃。如果我删除这行代码，它工作正常，但我添加它时崩溃。有人可以帮助我解释为什么这会崩溃。这是代码：

void assemblePass1(FILE * sourceF)
{
    int i;
    char line[260];
    char ** brokenUpLine;
    op * opCodes = createOPtable();     
    if (opCodes == NULL)
    {
        printf("Error allocating memory!\n\n");
        return;
    }

    char * fn = "intermediate.txt";
    FILE * imfp = checkForFileErrors(fn);
    if (imfp != NULL)
        imfp = fopen(fn, "w+");
    else
        return;                             

    while (fgets(line, 260, sourceF) != NULL)       //read line from source file
    {
        //write current source line to intermediate file
        fprintf(imfp, line);
        //breaks up line into values, write each value on new line in intermediate.txt
        brokenUpLine = breakUpLine(line);               //remember to free brokenUpLine
        strcat(brokenUpLine[0], "\n");
        strcat(brokenUpLine[1], "\n");
        strcat(brokenUpLine[2], "\n");
        strcat(brokenUpLine[4], "\n");          //*crashes at this line*
        fprintf(imfp, brokenUpLine[0]);         //write label to file
        fprintf(imfp, brokenUpLine[1]);         //write opcode to file
        fprintf(imfp, brokenUpLine[2]);         //write operand to file
        fprintf(imfp, brokenUpLine[3]);         //write comments to file
        fprintf(imfp, brokenUpLine[4]);         //write error code to file

        for (i = 0; i < 5; i++)
        {
            brokenUpLine[i] = NULL;
            free(brokenUpLine[i]);
        }

        free(brokenUpLine);
        break;              //for testing
    }

    free(opCodes);
    fclose(imfp);
    printf("Opening file was a success!\n\n");
}

这是breakUpLine（）：

char ** breakUpLine(char * line)
{
    int row = 5, cols = 150, i;
    char * string = NULL;
    char ** arr = (char**)malloc(row * sizeof(char*));  //allocate 5 rows of memory (label(0), opcode(1), operand(2), comment(3), error code(4))
    for (i = 0; i < row; i++)
        arr[i] = (char*)malloc(cols * sizeof(char));    //allocates 150 columns to store a string per row

    char * token = NULL;
    i = 0;

    if (line[0] == ' ')                 //if there is no label, assign to the label row an empty space
    {
        arr[0] = " ";                   //store a space in label column to indicate it has no label
        i = 1;
    }

    token = strtok(line, " \t");        //get first token; break up by spaces or tabs
    if (token[0] == '.')                    //if a period is found at beginning of line, then the line is an entire comment
    {
        token = strtok(NULL, "\n");     //store entire comment sentence including the spaces, so update token delimeter to a newline character
        arr[0] = ".";
        arr[1] = "\n";
        arr[2] = "\n";
        arr[3] = token;                 //store comment in comment row
        arr[4] = NULL;                  //error code will be 0 or NULL if no error was detected
        return arr;                     //remember to free arr in assemblePass1()
    }

    arr[i++] = token;                   //store token in appropriate row
    while (token != NULL && i != 3)     
    {
        token = strtok(NULL, " \t");
        arr[i++] = token;
        //totest:printf("%s\n", arr[i-2]);
    }

    //totest:printf("%s\n", arr[i - 1]);
    token = strtok(NULL, "\t");     //process comments
    arr[i] = token;                 //store comment in comment column
    arr[4] = "Error code";                  //error code will go here

    return arr;
}

Answer 1

因为这行而崩溃：

    arr[4] = NULL;                  //error code will be 0 or NULL if no error was detected

因为您随后尝试strcat指向此NULL指针：

    strcat(brokenUpLine[4], "\n");          //*crashes at this line*

也许这应该是：

    strcat(brokenUpLine[3], "\n");

???

你这里也有问题：

    arr[0] = ".";
    arr[1] = "\n";
    arr[2] = "\n";

这些应该是：

    strcpy(arr[0], ".");
    strcpy(arr[1], "\n");
    strcpy(arr[2], "\n");

否则，当您尝试strcat这些字符串时，（a）未定义的行为，（b）三次内存泄漏，（c）如果您稍后尝试free()这些字符串，则会有更多未定义的行为

尝试将换行符添加到char数组的末尾时，strcat不起作用

1 个答案: