c - 多维指针数组重新分配导致段错误

Question

首先，我将解释为什么我要这样做。我正在学习计算机编程类(class)，我的教授给我们布置了一项作业，我们必须制作一组记录(每个记录都包含名字、姓氏和分数)，然后允许用户使用以下命令操作记录菜单选项。所有这些都必须仅使用指针数组来完成，并且不允许使用结构。我知道这很令人头痛。我知道这可能是实现这一目标的最困难的方法之一，但这正是教授想要的。

除此之外，下面是到目前为止我的主要功能。大多数长 printf 函数只是我打印调试信息。请注意 char*** 变量的声明。它旨在充当 3D 数组，其中 nameRecords[0] 为第一条记录，nameRecords[0][0] 为第一条记录的名字，nameRecords[0][1] 是第一条记录的姓氏。第三个维度是 nameRecords[0][0][21]，因为字符串的长度仅为 20 个字符加上空字符。

int main(void)
{
    char ***nameRecords = NULL;
    float *scores = NULL;
    int size = 0; // total number of records
    int usrInt = 0;
    while(usrInt < 1)
    {
        printf("\nEnter the number of records to record(min 1):  ");
        scanf("%d", &usrInt);
        inpurge();

        if(usrInt < 1) printf("\nMust be integer greater than 1.\n");
    }

    nameRecords = (char***)calloc((size), sizeof(char**));
    scores = (float*)calloc(size, sizeof(float));

    int i;
    for(i = 0; i < usrInt; i++)
    {
        addRecord(&nameRecords, &scores, &size);
        printf("\nnameRecords@%p :: nameRecords[%d]@%p :: nameRecords[%d][0]=%s :: nameRecords[%d][1]=%s\n", nameRecords, size - 1, nameRecords[size - 1], size - 1, nameRecords[size - 1][0], size - 1, nameRecords[size - 1][1]);
    }

    printf("\nnameRecords[0]@%p\n", nameRecords[0]);

    prntRecords(nameRecords, scores, size);

    printf("\n\n\n");
    return 0;
}

在我第二次将&nameRecords传递到addRecord函数(定义如下)之后，麻烦就来了。澄清一下，如果用户选择在主函数开头仅输入 1 个条目，则不会收到段错误，并且程序实际上按预期运行和终止。

void addRecord(char ****records, float **scores, int *size)
{   
    printf("\t(*records)[0]%p\n", (*records)[0]);
    ++*size; // increment total number of records by 1
    int index = (*size) - 1;

    char ***tempNames = (char***)realloc(*records, (*size) * sizeof(char**)); // reallocate larger space.
    if(tempNames != *records)
        *records = tempNames; // set original pointer to new value.

    printf("\n\tsize - 1 = %d\n", index);

    float *tempScores = (float*)realloc(*scores, (*size) * sizeof(float)); // reallocate larger space.
    if(tempScores != *scores)
        *scores = tempScores; // set original pointer to new value.

    printf("\ttempNames[0]@%p\n", tempNames[0]);

    tempNames[index] = (char**)calloc(tempNames[index], 2 * sizeof(char*));
    enterRecord(tempNames[index], scores[index]);

    printf("\n\ttempNames@%p :: tempNames[0]@%p :: tempNames[%d][0]=%s :: tempNames[%d][1]=%s\n", tempNames, tempNames[0], index, tempNames[index][0], index, tempNames[index][1]);
    printf("\n\t*records@%p :: *records[0]@%p :: *records[%d][0]=%s :: *records[%d][1]=%s\n", *records, (*records)[0], index, (*records)[index][0], index, (*records)[index][1]);

    return;
}

下面是该程序的示例输出。不用花太多时间来解释发生了什么，选项卡式行是 addRecord 函数内的输出行。具体来说，在第二次通过 addRecord 函数时，指向第一条记录 record[0] 的指针已变成垃圾值，就在 EnterRecord 函数。

Enter the number of records to record(min 5):  2
        (*records)[0](nil)

        size - 1 = 0
        tempNames[0]@(nil)

Enter first name: 1

Enter last name: 1

Enter score: 1

COMPLETE enterRecord

        tempNames@0x6387010 :: tempNames[0]@0x6387050 :: tempNames[0][0]=1 :: tempNames[0][1]=1

        *records@0x6387010 :: *records[0]@0x6387050 :: *records[0][0]=1 :: *records[0][1]=1

nameRecords@0x6387010 :: nameRecords[0]@0x6387050 :: nameRecords[0][0]=1 :: nameRecords[0][1]=1
        (*records)[0]0x6387050

        size - 1 = 1
        tempNames[0]@0x6387050

Enter first name: 2

Enter last name: 2

Enter score: 2

COMPLETE enterRecord

        tempNames@0x6387010 :: tempNames[0]@0x40000000 :: tempNames[1][0]=2 :: tempNames[1][1]=2

        *records@0x6387010 :: *records[0]@0x40000000 :: *records[1][0]=2 :: *records[1][1]=2

nameRecords@0x6387010 :: nameRecords[1]@0x63870b0 :: nameRecords[1][0]=2 :: nameRecords[1][1]=2

nameRecords[0]@0x40000000

records@0x6387010 :: records[0]@0x40000000
Segmentation fault

所有调试信息都指向 enterRecord 函数是罪魁祸首。这就是邪恶的 enterRecord 函数...

void enterRecord(char **names, float *score)
{
    names[0] = (char*)calloc(21, sizeof(char)); // allocate first name string
    names[1] = (char*)calloc(21, sizeof(char)); // allocate last name string

    printf("\nEnter first name: ");
    fgets(names[0], 21, stdin);
    if(strlen(names[0]) == 20) // IGNORE. just handles overflow from fgets.
        inpurge();
    remNewLine(names[0]); // removes '\n' character at end of string

    printf("\nEnter last name: ");
    fgets(names[1], 21, stdin);
    if(strlen(names[1]) == 20) // IGNORE. just handles overflow from fgets.
        inpurge();
    remNewLine(names[1]); // removes '\n' character at end of string

    printf("\nEnter score: ");
    scanf("%f", score);
    inpurge();

    printf("\nCOMPLETE enterRecord\n");
    return;
}

只是...没有尝试更改受影响的指针。记录数组第二个元素的指针值(records[1])被传递到函数中，我看不到任何内容正在改变记录数组第一个元素的指针值(records[0])，但 records[0] 的值是导致段错误的原因。

对于代码的长度和所有混淆性的内容，我深表歉意。同样，这似乎是编写该程序的一种糟糕方法，但这正是情况所需要的。我只是为那位可怜的助教感到难过，他必须给其中 30 多份作业评分。

欢迎任何帮助。

Answer 1

这个问题似乎更好地实现为

#define MAX_FIRST_NAME_LEN (21)
#define MAX_LAST_NAME_LEN  (21)
#define MAX_SCORES         (10)

// in file global memory...
static char **ppFirstNames = NULL; 
static char **ppLastName   = NULL;
static int  **ppScores     = NULL;
static int    numOfEntries = 0;

// in the record input function, which needs NO parameters
scanf ( "%d", &numOfEntries );
if scanf fails, exit

ppFirstNames = malloc (numOfEntries*sizeof char*);
if malloc fails, exit

memset (ppFirstName, '\0', numOfEntries*sizeof char* );

ppLastName = malloc (numOfEntries*sizeof char*);
if malloc fails, free all, exit

memset (ppLastName, '\0', numOfEntries*sizeof char* );

ppScores = malloc (numOfEntries *sizeof int* );
if malloc fails, free all, exit

for(int i=0; i<numOfEntries; i++ )
    ppFirstNames[i] = malloc( MAX_FIRST_NAME_LEN );
    if malloc fails free all, exit

    memset ( ppFirstNames[i], '\0', MAX_FIRST_NAME_LEN );

    ppLastName[i] = malloc (MAX_LAST_NAME_LEN);
    if malloc fails free all, exit

    memset ( ppLastName[i], '\0', MAX_LAST_NAME_LEN );

    ppScores[i] = malloc (MAX_SCORES *sizeof int);-1
    if malloc fails, free all, exit

    memset (ppScores[i], '\0', MAX_SCORES *sizeof int );
end for

for ( int i=0; i < numOfEntries; i++ )
    now read each record
    scanf( "%(MAX_FIRST_NAME_LEN-1)s", ppFirstNames[i] );
    if scanf fails, free all, exit

    scanf( "%(MAX_LAST_NAME_LEN-1)s", ppLastNames[i] );
    if scanf fails, free all exit

    for( int j=0; j< MAX_SCORES; j++ )
        now read this students scores
        int tempScore;  
        scanf( "%d", tempScore );
        if scanf fails, free all, exit

        if -1 == tempScore ) break;

        ppScores[i][j] = tempScore;
    end for
end for

以上是输入记录的伪代码

并且应该足以使输入正确。

此后打印信息应该很容易。