[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A和B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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