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指向結構引用的指針數組

[英]Array of Pointers to Structs referencing

我正在編寫“納米社交網絡”,僅通過注冊成員並使用結構體中的指針將它們連接即可。 我正在嘗試使用connect函數,但不確定在其中是否以正確的方式編寫它,因為當我嘗試運行該函數時該程序崩潰,並且收到0錯誤和0警告(使用CodeBlocks進行高級警告)。

首先,我注冊了至少2個成員,然后當我嘗試通過connect()函數將它們連接時,我發現我想連接的用戶不存在,這是該函數中條件的一部分。 是我注冊的用戶沒有保存在內存中嗎? 還是我寫錯了什么?

可以通過各種方式簡化connect函數,但是一旦我弄清了自己的錯誤,我就會這樣做。

編碼:

#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,當我至少注冊了3個成員並輸入現有名稱后使用connect函數時,我總是得到答案“您輸入的用戶不存在”,這是我沒有將其保存在內存中還是我缺少功能嗎?

您不能像這樣比較char數組,您要比較這些指針指向的地址:

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

您必須使用strcmp函數:

strcmp(ptr[i]->name, ptr[0]->name)

另外,考慮一下您的connect函數-為什么使用第一個member作為輸入scanf("%s", ptr[i]->name); 您寧願鍵入成員AB的名稱,在ptr數組中找到它們的位置,然后連接它們。

我也不明白為什么在member本身中有這五個指向member指針。 您已經有一個數組。

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