指向结构的指针数组

时间:2015-12-20 14:05:08

标签: c arrays pointers struct

我正在编写一个“纳米社交网络”,只需注册一个成员并使用结构中的指针连接它们。我正在尝试使用一个连接函数,其中我不确定我是以正确的方式编写它,因为当我尝试运行它时程序崩溃并且我得到0个错误和0个警告(使用CodeBlocks在高级警告)。

首先我注册了至少2个成员然后当我尝试通过connect()函数连接它们时,我得到了我想要连接的用户不存在,这是函数中条件的一部分。我正在注册的用户是不是被保存在内存中?或者我写错了什么?

连接功能可以通过各种方式简化,但是一旦我清楚自己的错误,我就会这样做。

代码:

#include <stdio.h>
#include <stdlib.h>

struct member{
    int num;
    char name[10];
    struct member *m1;
    struct member *m2;
    struct member *m3;
    struct member *m4;
    struct member *m5;
}*ptr[5];

void addmember(){
    int i;

    printf("Enter new members details:\n");
        for(i=0; i<4; i++){
        ptr[i] = malloc(sizeof(struct member));
        printf("\n Enter ID number:\n");
        scanf("%d", &ptr[i]->num);
        printf("\n Enter Name:\n");
        scanf("%s", ptr[i]->name);
    }

    printf("\n Added member details are:");
        for(i=0; i<4; i++){
        printf("\n ID number : %d", ptr[i]->num);
        printf("\nName : %s", ptr[i]->name);
        }
}
void connect(){ //when always typing the first username
    int i=0;
    printf("Please type in your user name : \n");
    scanf("%s", ptr[i]->name);

    if(ptr[i]->name == ptr[0]->name){ //being first user connecting.
        printf("Please type existing member name you wish to connect with :\n");
        scanf("%s", ptr[i]->name);

        if(ptr[i]->name == ptr[1]->name){
            ptr[0]->m1 = ptr[1]->m1;
            ptr[1]->m2 = ptr[0]->m2;
            printf("member 1 connected with member 2!\n");
        }
        else if(ptr[i]->name == ptr[2]->name){
            ptr[0]->m1 = ptr[2]->m1;
            ptr[2]->m3 = ptr[0]->m3;
            printf("member 1 connected with member 3!\n");
        }
        else if(ptr[i]->name == ptr[3]->name){
            ptr[0]->m1 = ptr[3]->m1;
            ptr[3]->m4 = ptr[0]->m4;
            printf("member 1 connected with member 4!\n");
        }
        else if(ptr[i]->name == ptr[4]->name){
            ptr[0]->m1 = ptr[4]->m1;
            ptr[4]->m5 = ptr[0]->m5;
            printf("member 1 connected with member 5!\n");
        }
         else{
            printf("User you typed in does not exist\n");
        }
    }

}

int main(){
    int op = 0;

    printf("Welcome to NSN, please register 4 members to proceed to options menu.\n");
    addmember(); //adding 4 members

    //general member section
    printf("Now please select an option by typing option number\n");
    while(op != 3){
        printf("\n");
        printf("1. Add new member.\n 2. Connect with member.\n 3. Exit\n");
        printf("Enter your choice:\n");
        scanf("%d", &op);
        switch(op){
        case 1:
             addmember();
            break;
        case 2:
            connect();
            break;
        case 3:
             printf("Bye!\n");
            exit(0);
             break;
         default:
             printf("Invalid choice!\n");
         }
    }
    return(0);
 }

如前所述,当我在注册至少3名成员并输入现有名称后使用连接功能时,我总是得到答案“你输入的用户不存在”,这是某种方式,我不是在保存这个记忆还是我错过了一个功能?

1 个答案:

答案 0 :(得分:2)

你无法比较像这样的scope_master数组,你要比较这些指针指向的地址:

char

您必须使用strcmp功能:

ptr[i]->name == ptr[0]->name

另外,请考虑一下您的strcmp(ptr[i]->name, ptr[0]->name) 功能 - 为什么使用第一个connect输入 - member?您更愿意输入成员 A B 的名称,在scanf("%s", ptr[i]->name);数组中找到他们的位置,然后连接它们。

我也不明白为什么你在ptr本身有member的这五个指针。你已经有了一个数组。