realloc指针数组无所事事

Question

我有一个字符串数组,当它不再有NULL指针(意味着数组已满)时,我想要它.我已经尝试了realloc并没有成功,我想我并没有正确地指向思考.

这是我的代码:

int storage; //global, outside of main
int i, key;
char **people;
char **phones;

printf("Please enter a storage cacpity:\n");
scanf("%d",&storage);
printf("\n");

people=malloc(storage*sizeof(char *));
phones=malloc(storage*sizeof(char *));

for (i=0; i<storage; i++) {
    people[i] = NULL;
    phones[i] = NULL;
}

void AddNewContact(char * people[], char * phones[]) {
    char name[100];
    char phone[12];
    int i, listfull = 0;

    printf("Enter a contact name:\n");
    scanf("%s",&name);
    printf("Enter a phone number:\n");
    scanf("%s",&phone);

    for (i=0; i<storage; i++) {
        if (people[i]==NULL) {
            people[i] = (char *)malloc(strlen(name));
            phones[i] = (char *)malloc(strlen(phone));
            strcpy(people[i],name);
            strcpy(phones[i],phone);
            break;
        }
        listfull = 1;
    }

    if (listfull == 1) {
        storage++;
        people = realloc(&people,(storage)*sizeof(char *));
        phones = realloc(&phones,(storage)*sizeof(char *));
        people[storage-1] = NULL;
        phones[storage-1] = NULL;
        strcpy(people[storage-1],name);
        printf("\nData Base extanded to %d",storage);
    }
    printf("\n");
    return;
}

void PrintAll(char * people[], char * phones[]) {
    int i;
    for (i=0; i<storage; i++) {
        if (NULL != people[i]) {
            printf("Name: %s, ",people[i]);
            printf("Number: %s\n",phones[i]);
        }
    }
    printf("\n");
    return;
}

任何帮助都将受到高度赞赏,我坚持了几个小时,没有运气这样做.

Answer 1

你有4个重要的错误,首先你将数组的地址传递给scanf()那个错误,你应该改变

scanf("%s", &name);

至

scanf("%s", name);

另外scanf("%s",&phone);,我还应该建议使用长度说明符scanf来防止缓冲区溢出

scanf("%99s", name);

即终止符的name数组-1 的长度'\0'.

其次,你的realloc调用也是错误的,你应该传递指针而不是它的地址,而不是这个

people = realloc(&people,(storage)*sizeof(char *));

你应该做这个

people = realloc(people, storage * sizeof(char *));

但即使这不是百分之百正确,因为如果realloc失败,你将覆盖指针,然后你就没有机会清理内存,所以你应该做一些像

void *pointer;
pointer = realloc(people, storage * sizeof(char *));
if (pointer == NULL)
    free_people_andCleanUpOtherResourcesAndExitFromThisFunction();
people = pointer;

同样的道理phones.

第三,你应该总是为一个额外的角色分配空间,终止'\0',这个

people[i] = (char *)malloc(strlen(name));

应该读

people[i] = malloc(1 + strlen(name));

请注意我删除了不必要的演员表.

第四,你在第一次迭代中离开循环listfull == 1,即使有列表还没有完整.

for (i=0; i<storage; i++) {
    if (people[i]==NULL) {
        people[i] = malloc(1 + strlen(name));
        phones[i] = malloc(1 + strlen(phone));
        strcpy(people[i],name);
        strcpy(phones[i],phone);
        break;
    }
    listfull = 1;
}

我建议在循环之外

listfull = (i == storage);

注意:函数失效的可能性并不重要,如果它理论上会失败,你应该总是检查它是否失败,这样可以节省你几个小时的调试时间来找到一个非常愚蠢的错误,你没有检查是否可能失败.