c - 使用动态内存分配 i C 时出现奇怪的输出

Question

我正在编写这段代码，但尚未完成。但是在测试添加新联系人功能和打印功能时，我得到了这个奇怪的输出。只要我不添加多个联系人，一切看起来都很好。我在这里做错了什么？

//Libraries:
#include <stdio.h>
#include <stdlib.h>

//Function prototypes
int showMenu(); //Shows user options

//Main function
int main() {

    //Variable declerations
    int i = 0;
    int iOption = 0;
    int iContacts = 0;
    char *cPhoneBook_name;
    int *iPhoneBook_nr;

    cPhoneBook_name = (char *) calloc(1, 80 * sizeof(char));
    iPhoneBook_nr = (int *) calloc(1, sizeof(int));

    do {

        //Memory check
        if(cPhoneBook_name == NULL || iPhoneBook_nr == NULL){
            printf("\nOut of Memory!\n");
            return;
        }

        //Show menu
        iOption = showMenu();

        switch (iOption) {
            case 1: //Add contact
                printf("\nEnter name:\n");
                scanf("%s", &cPhoneBook_name[iContacts]);
                printf("\nEnter number:\n");
                scanf("%d", &iPhoneBook_nr[iContacts]);
                cPhoneBook_name = realloc(cPhoneBook_name, 80 * sizeof(char));
                iPhoneBook_nr = realloc(iPhoneBook_nr, sizeof(int));
                iContacts += 1;
                break;
            case 2: //Modify contact
                //Code here
                break;
            case 3: //Show contacts
                for(i = 0 ; i < iContacts ; i++) {
                    printf("\n%s %d\n", &cPhoneBook_name[i], iPhoneBook_nr[i]);
                }
                break;
            case 4: //Free memory (delete contacts)
                free(cPhoneBook_name);
                free(iPhoneBook_nr);
                break;
        }
    }while(iOption != 5);//Exit if iOption is 5;

    system("clear");
    return 0;
}


//Function definition - showMenu()
int showMenu() {

    int iOption = 0;

    //system("clear");

    printf("\n   Phone book   \n");
    printf("------------------------\n");
    printf("1. Add new contact.\n");
    printf("2. Edit existing contact\n");
    printf("3. Show contact(s)\n");
    printf("4. Clear phone book\n");
    printf("5. Exit\n");
    printf("------------------------\n");
    printf("Option --> ");
    scanf("%d", &iOption);

    return iOption;
}
//---------------------------------------------------------

Answer 1

问题在于:

printf("\n%s %d\n", &cPhoneBook_name[i], iPhoneBook_nr[i]);

cPhoneBook_name 如果类型为 char[]。因此，例如表达式 char[1] 为您提供列表中的第二个字符。虽然代码中的名称是 80 个 char 宽。所以尝试:

printf("\n%s %d\n", &cPhoneBook_name[i*80], iPhoneBook_nr[i]);

编辑:

我还发现:

cPhoneBook_name = realloc(cPhoneBook_name, 80 * sizeof(char));

没有效果，因为您正在将 cPhoneBook_name 的大小从 80 调整到 80。尝试:

    cPhoneBook_name = realloc(cPhoneBook_name, (iContacts+1) 80 * sizeof(char)); 

还有: scanf("%s", &cPhoneBook_name[80*iContacts]);