尝试在结构列表中打印元素的结构向量

Question

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

const int n = 20;

定义结构体和列表

typedef struct pole{
   int year;
   char county[n];
   char ID_election[n];
   char ID_electionUser[n];
   int num_candidates;
   char name_candidate[n]; 
   int limit_votes;
}election;

typedef struct polelist{
   election voting;
   struct polelist *next_pole;
}node_pole;

功能

void createElection(node_pole *head_pole, election *info);
void insertElection(node_pole **head_pole, election *info);
void printElection(node_pole *head_pole);

主要

int main (){

node_pole *head_pole = NULL;
election *info = (election*)malloc(sizeof(election));
int i = 0;

我创建选举，然后插入节点。 我打印了刚刚创建的选举并且它起作用了（我可以读到候选人的名字）。

createElection(head_pole, info);
insertElection(&head_pole, info);
printf("\nDo you want to visualize your election? Y/N: ");
printf("\n----- Last election created -----\n ");
printf("\nID Election:%s  \nID Elect_user:%s  \nYear:%d\nCounty:%s  \nMax voters:%d", info->ID_election, info->ID_electionUser, info->year, info->county, info->limit_votes); 
printf("\nCandidates' number:%d", info->num_candidates);
for(i = 0; i < info->num_candidates; i++){
    printf("\nCandidates' name [%d]: %s", i+1, ((info + i)->name_candidate));
}

这是终端上的结果

ID Election:D1  
ID Elect_user:S1  
Year:2023 
County:USA  
Max voters:5
Candidates' number:3
Candidates' name [1]: A1
Candidates' name [2]: B2
Candidates' name [3]: C3

我再次打印选举结果，但使用函数 printElection。 我在这里遇到了问题。我看不清候选人的名字，只能读到第一个。

printf("\n\t\t\t-ALL-\n\n");    
printElection(head_pole); 

return 0;
}

通过终端中的功能打印选择：

ID Election:D1 
ID Elect_User: S1  
Year:2023 
County:USA  
Max voters:5
Candidates' number:3
Candidates' name [1]: A1
Candidates' name [2]: 
Candidates' name [3]:

功能定义

void createElection(node_pole *head_pole, election *info){

        election *pole = info;    
        int i = 0;

        printf("\nInsert data:  \n Year: ");
        scanf("%d", &pole->year);
        printf(" County:");
        scanf("%s", pole->county);
        printf(" Voters' limit:");
        scanf("%d", &pole->limit_votes);  
        printf("ID_Election: ");         
        scanf("%s", pole->ID_election);
        printf("ID_Election_User: ");         
        scanf("%s", pole->ID_electionUser);
        printf("Candidates' number: ");
        scanf("%d", &pole->num_candidates);
            for(i = 0; i < pole->num_candidates; i++){  
                    printf("\n Candidates' names [%d]:", i+1);
                    scanf("%s", (pole+i)->name_candidate);
            }                
return;
}

void insertElection(node_pole **head_pole, election *info){

        node_pole *newnode = (node_pole*)malloc(sizeof(node_pole));
       
        newnode->voting = *info;
        newnode->next_pole = *head_pole;
        *head_pole = newnode;
return;
}

void printElection(node_pole *head_pole){

    int i;
    node_pole *tmp = head_pole;
    
    if(tmp == NULL){
            printf("\nNo poll in the database.");
            return;
    }
        while(tmp != NULL){
            printf("\n----- Your election ----- ");
            printf("\nID Election:%s \nID Elect_User: %s  \nYear:%d \nCounty:%s  \nMax voters:%d", tmp->voting.ID_election, tmp->voting.ID_electionUser, tmp->voting.year, tmp->voting.county, tmp->voting.limit_votes);
            printf("\nCandidates' number:%d", tmp->voting.num_candidates);
            for(i = 0; i < tmp->voting.num_candidates; i++){
                printf("\nCandidates' name [%d]: %s", i + 1, (tmp+i)->voting.name_candidate);
            }                    
            tmp = tmp->next_pole;
        }
return;
}

Answer 1

正如评论中所指出的，目前选举中只能容纳一个候选人姓名（20 - 1 个字符长）。

这是错误的：

printf("\nCandidates' name [%d]: %s", i+1, ((info + i)->name_candidate));

像上面这样的引用将

info

视为数组，并且等同于

info[i]->name_candidate

。

info

不是数组。

推动这一进程的小改变（并希望增加OP的理解）：

#define MAX_CANDIDATES 8 // an arbitrary number

typedef struct pole{
   int year;
   char county[n];
   char ID_election[n];
   char ID_electionUser[n];
   int num_candidates;
   char name_candidate[MAX_CANDIDATES][n]; // 2D array to hold up to 8 x 20 characters
   int limit_votes;
} election;

和

        printf("# of Candidates: ");
        scanf("%d", &pole->num_candidates);

        if( pole->num_candidates > MAX_CANDIDATES )
            // handle erroneous entry

        for(i = 0; i < pole->num_candidates; i++){  
            printf("\n Candidate's name [%d]:", i+1);
            scanf("%19s", pole->name_candidate[i] );
        }                

并修复打印语句：

printf("Candidate name [%d]: %s\n", i+1, info->name_candidate[i]);

这些变化应该能让OP取得一些成功并推动项目向前发展。值得重新审视这一点，既不要为较少的候选人使用过多的空间，也不要限制 8 名候选人。这是留给读者的练习。

最后：每次调用

scanf()

都会返回一个当前被忽略的值。获取用户输入充满挑战。为读者提供的另一个练习，以提高此代码的稳健性。