家庭成员结构,C

时间:2018-04-22 16:08:12

标签: c struct scanf

我是学生,我们的任务之一就是在C上编写一个有两种结构的程序:一个用于家庭,一个用于人(如下所示)。 我们应该获取包括姓名和ID在内的家庭成员信息,然后使用名称字段的动态分配(在人员结构中)打印它。

继承我的代码,但是它出了问题。我不会让我继续处理母亲的详细信息......我的意思是,它会跳过你需要插入名称的部分,直接进入& #34; ID&#34 ;.  我试图添加" flushall"但似乎没有用。可能还有更多我不知道的代码问题,因为我还在试图找出这个问题的错误:

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

typedef struct Person {
    char* name;
    long id;
}person;

typedef struct Family {
    person dad;
    person mom;
    int numOfKids;
    person* pointers[10];
}family;

void readDate(person* t);
void printData(person* t);
void readFamily(family* f);
void printFamily(family* f);

void main()
{ 
    family ff;
    readFamily(&ff);
    printFamily(&ff);
    free(ff.dad.name);
    free(ff.mom.name);
    for (int i = 0; i < ff.numOfKids; i++)
        free(ff.pointers[i]->name);

}
void readDate(person* t)
{
    int size;
    char names[200];
    printf("Please enter a name\n");
    _flushall;
    gets_s(names);
    size = strlen(names);
    t->name = (char*)calloc(1, sizeof(names + 1));
    t->name = names; 
    printf("Please enter ID\n");
    scanf("%d", &t->id);
}
void printData(person* t)
{
    printf("%s %d\n", t->name, t->id);
}
void readFamily(family* f)
{
    (f->numOfKids) = 0;
    int countinue = 1;
    char c;
    printf("Please enter the father's info\n");
    _flushall;
    readDate(&(f->dad));
    printf("Please enter the mother's info\n");
    _flushall;
    readDate(&(f->mom));
    do {
        printf("Would u like to add a child? Y/N\n");
        _flushall;
        scanf("%c", &c);
        if (c == 'N' || c == 'n')
        {
            countinue = 0;
        }
        else
        {
            printf("Pleae enter the child's info:\n");
            readDate((f->pointers[f->numOfKids]));
            f->numOfKids++;
        }
    } while (countinue &&f->numOfKids <= 10);
}
void printFamily(family* f)
{
    printf("The family's parents info:\n");
    printf("Mom:%s ID:%d\t Dad:%s ID:%d\n", f->mom.name, f->mom.id, f->dad.name, f->dad.id);
    if (f->numOfKids == 0)
        printf("There are no children!\n");
    else
    {
        for (int i = 0; i < f->numOfKids; i++)
        {
            printf("#%d:Name:%s ID:%d\n", i + 1, f->pointers[i]->name, f->pointers[i]->id);
        }
    }
    printf("\n");
}

1 个答案:

答案 0 :(得分:0)

在您的代码中使用rewind(stdin)代替flushall,因为flushall在VS中是过时的功能