c - C 语言的基本结构帮助

Question

如果用户尝试搜索先前未在结构中输入的名称，我需要使第二个函数打印“没有具有该名称的条目”。除了这部分之外，一切都工作正常。我当前的 else 语句有效，但它每次都会输出该行，而不仅仅是在结构中未输入名称时。有什么帮助可以改变吗？

#include <stdio.h>
#include <string.h>
#include "libpb.h"

void add_person(struct phone_book * pb, struct personal_info person)
{
    int num = pb->num_people;
    strcpy(pb->person[num].first, person.first);
    strcpy(pb->person[num].last, person.last);
    strcpy(pb->person[num].phone, person.phone);
    num++;
    pb->num_people = num;
}

void search_pb(struct phone_book pb, char find_name[])
{
    int p;
    for (p = 0; p < pb.num_people; p++)
    {
        if (strcmp(find_name, pb.person[p].first) == 0)
        {
            printf("\nName: %s %s\n", pb.person[p].first, 
            pb.person[p].last);
            printf("Phone: %s\n", pb.person[p].phone);
        } else
        {
            printf("No entries with that name. \n");
        }
    }
}

我得到了要使用的主函数phone_book.c，所以我只需要编写上面的函数和一个头文件:

int main () 

{

char cont;
char find_name[25];
struct phone_book pb;
pb.num_people = 0;
struct personal_info person;

printf("\n*********************************************\n");
printf("\n      Start with entering new contacts!      \n");
printf("\n*********************************************\n");
printf("\nWould you like to enter a new contact (Y/N): ");

while(pb.num_people < 20) 
{
    scanf("%c", &cont);

    if (cont == 'Y') 

    {
        printf("Enter a first name: ");
        scanf("%s", person.first);
        printf("Enter %s's last name: ", person.first);
        scanf("%s", person.last);
        printf("Enter %s's phone number: ", person.first);
        scanf("%s", person.phone);
        add_person(&pb, person);
    }

    else if (cont == 'N') break;
    else if (cont == '\n') continue;
    else printf("Error: User entered '%c'. Must enter either 'Y' or 'N'\n", 
    cont);

    printf("\nWould you like to enter a new name (Y/N): ");

}

//search phone book by first name and print persons

printf("\n*********************************************\n");
printf("\n        Now You can search for names!        \n");
printf("\n*********************************************\n");
printf("\nWould you like to search for a name (Y/N)? ");

while(1)
{
    scanf("%c", &cont);

    if (cont == 'Y')
    {
        printf("Enter a person's name to search for: ");
        scanf("%s", find_name);
        //scanf("%c", &tmp);
        search_pb(pb, find_name);
    }

    else if (cont == 'N') break;
    else if (cont == '\n') continue;
    else printf("Error: User entered '%c'. Must enter either 'Y' or 'N'\n", 
    cont);

    printf("\nWould you like to search for a name (Y/N)? ");

}
return 0;
}

我也已经制作了必要的头文件 libpb.h:

#include<stdio.h>
#include<string.h>
#define MAX 20
#define _CRT_SECURE_NO_DEPRECATE


struct personal_info 

{

char first[25];

char last[25];

char phone[15];

};

struct phone_book 

{

struct personal_info person[MAX];

int num_people;

};

void add_person(struct phone_book *pb, struct personal_info person);

void search_pb(struct phone_book pb, char find_name[]);

Answer 1

所以我在 repl.it ( https://repl.it/repls/SphericalFuchsiaIntroductory ) 中输入了您的代码并声明缩小问题范围。根据我的测试，只有当您在结构中输入 2 个或更多名称时，才会发生此问题，这使我相信该函数被调用了两次或其他原因。

您的代码的问题是 for 循环内的 else 语句，您的循环遍历所有联系人并搜索匹配的联系人，如果只有一个联系人，则 else 会正常触发，但如果有多个联系人它将触发与循环一样多的次数。为了在 repl.it 上修复它(如果您想查看完整内容，请单击链接)，我将您的函数更改为:

void search_pb(struct phone_book pb, char find_name[]) {
    int p;

    for (p = 0; p < pb.num_people; p++)
    {
        if (strcmp(find_name, pb.person[p].first) == 0)
        {
            printf("\nName: %s %s\n", pb.person[p].first, 
                pb.person[p].last);

            printf("Phone: %s\n", pb.person[p].phone);
            return;
        }

    }
    printf("No entries with that name. \n");
}

如您所见，现在您的函数一旦找到匹配项就会返回并打印匹配项，否则它默认打印未找到的条目。 希望有帮助!