关于#c语言#的问题，请各位专家解答！
这个代码应该如何改正，只要选则已婚数字2或者离婚数字三，输出的时候都会只显示未婚，求改正方法（为什么输出函数中switch中默认为一选项，还是那个变量一直是一，可是我在输入函数中给那个变量赋值了啊？）
```c
#include<stdio.h>
#include<stdlib.h>
struct data
{
int year;
int month;
int day;
};
struct marriedstate
{
struct data marrday;
char spousename;
int children;
};
struct divorcestate
{
struct data divorcedata;
int children;
};
union marritalstate
{
int marryflay;
struct marriedstate married;
struct divorcestate divorce;
};
struct person
{
int no;
char name[10];
char sex[5];
int age;
union marritalstate marrital;
};
void _input(struct person *woker,int num);
void _output(struct person *woker,int num);
int main()
{
struct person *woker;
int num,i;
printf("请输入员工数：");
scanf("%d",&num);
woker=(struct person*)malloc(sizeof(struct person)*num);
_input(woker,num);
_output(woker,num);
return 0;
}
void _input(struct person *woker,int num)
{
int i,n=1;
for(i=0;i<num;i++)
{
printf("请输入第%2d个员工的姓名 性别 年龄：",i+1);
woker[i].no=i+1;
scanf("%s %s %d",&woker[i].name,&woker[i].sex,&woker[i].age);
printf("请输入第%2d个员工的婚姻状态：\n",i+1);
printf("\t1.未婚\n\t2.已婚\n\t3.离婚\n");
do{
printf("请速回如第%2d个员工的婚姻状态对应的号（1-%d）:",i+1,num);
scanf("%d",&woker[i].marrital.marryflay);
}while(woker[i].marrital.marryflay<1||woker[i].marrital.marryflay>3);
switch(woker[i].marrital.marryflay)
{
case 1:break;
case 2:printf("请输入第%2d个员工的结婚纪念日：",i+1);
scanf("%d %d %d",&woker[i].marrital.married.marrday.year,&woker[i].marrital.married.marrday.month,&woker[i].marrital.married.marrday.day);
printf("请输入第%2d个员工的配偶姓名：",i+1);
scanf("%s",&woker[i].marrital.married.spousename);
printf("请输入第%2d个员工的孩子数：",i+1);
scanf("%d",&woker[i].marrital.married.children); break;
case 3:printf("请输入第%2d个员工的离婚年月日：",i+1);
scanf("%d %d %d",&woker[i].marrital.divorce.divorcedata.year,&woker[i].marrital.divorce.divorcedata.month,&woker[i].marrital.divorce.divorcedata.day);
printf("请输入第%2d个员工的孩子数：",i+1);
scanf("%d",&woker[i].marrital.divorce.children); break;
}
}
}
void _output(struct person *woker,int num)
{
printf("员工编号 姓名 性别 年龄 婚姻状态 结婚或离婚日期 配偶姓名 孩子数\n");
for(int i=0;i<num;i++)
{
printf("%4d%6s %3s %2d",woker[i].no,woker[i].name,woker[i].sex,woker[i].age);
switch(woker[i].marrital.marryflay)
{
case 1:printf("\t 未婚\n");break;
case 2:printf("\t 已婚");
printf("\t%4d/%2d/%2d",woker[i].marrital.married.marrday.year,woker[i].marrital.married.marrday.month,woker[i].marrital.married.marrday.day);
printf("%8s %2d\n",woker[i].marrital.married.spousename,woker[i].marrital.married.children ); break;
case 3:printf("\t 离婚");
printf("\t%4d/%2d/%2d",woker[i].marrital.divorce.divorcedata.year,woker[i].marrital.divorce.divorcedata.month,woker[i].marrital.divorce.divorcedata.day);
printf("\t\t%d\n",woker[i].marrital.divorce.children); break;
}
}
}
```# 我想要达到的结果，如果你需要快速回答，请尝试 “付费悬赏”
问题出在union marritalstate，联合体的元素公用同一个内存空间，输入完marryflay后再输入日期，marryflay的值会被覆盖。
注释处是修改的地方
#include <stdio.h>
#include <stdlib.h>
struct data
{
int year;
int month;
int day;
};
struct marriedstate
{
struct data marrday;
char spousename[10]; //
int children;
};
struct divorcestate
{
struct data divorcedata;
int children;
};
// union marritalstate
struct marritalstate
{
int marryflay;
struct marriedstate married;
struct divorcestate divorce;
};
struct person
{
int no;
char name[10];
char sex[5];
int age;
// union marritalstate marrital;
struct marritalstate marrital;
};
void _input(struct person *woker, int num);
void _output(struct person *woker, int num);
int main()
{
struct person *woker;
int num, i;
printf("请输入员工数：");
scanf("%d", &num);
woker = (struct person *)malloc(sizeof(struct person) * num);
_input(woker, num);
_output(woker, num);
return 0;
}
void _input(struct person *woker, int num)
{
int i, n = 1;
for (i = 0; i < num; i++)
{
printf("请输入第%2d个员工的姓名 性别 年龄：", i + 1);
woker[i].no = i + 1;
// scanf("%s %s %d", &woker[i].name, &woker[i].sex, &woker[i].age);
scanf("%s %s %d", woker[i].name, woker[i].sex, &woker[i].age);
printf("请输入第%2d个员工的婚姻状态：\n", i + 1);
printf("\t1.未婚\n\t2.已婚\n\t3.离婚\n");
do
{
printf("请速回如第%2d个员工的婚姻状态对应的号（1-%d）:", i + 1, num);
scanf("%d", &woker[i].marrital.marryflay);
} while (woker[i].marrital.marryflay < 1 || woker[i].marrital.marryflay > 3);
switch (woker[i].marrital.marryflay)
{
// case 1:
// break;
case 2:
printf("请输入第%2d个员工的结婚纪念日：", i + 1);
scanf("%d %d %d", &woker[i].marrital.married.marrday.year, &woker[i].marrital.married.marrday.month, &woker[i].marrital.married.marrday.day);
printf("请输入第%2d个员工的配偶姓名：", i + 1);
// scanf("%s", &woker[i].marrital.married.spousename);
scanf("%s", woker[i].marrital.married.spousename);
printf("请输入第%2d个员工的孩子数：", i + 1);
scanf("%d", &woker[i].marrital.married.children);
break;
case 3:
printf("请输入第%2d个员工的离婚年月日：", i + 1);
scanf("%d %d %d", &woker[i].marrital.divorce.divorcedata.year, &woker[i].marrital.divorce.divorcedata.month, &woker[i].marrital.divorce.divorcedata.day);
printf("请输入第%2d个员工的孩子数：", i + 1);
scanf("%d", &woker[i].marrital.divorce.children);
break;
}
}
}
void _output(struct person *woker, int num)
{
printf("员工编号 姓名 性别 年龄 婚姻状态 结婚或离婚日期 配偶姓名 孩子数\n");
for (int i = 0; i < num; i++)
{
printf("%4d%6s %3s %2d", woker[i].no, woker[i].name, woker[i].sex, woker[i].age);
switch (woker[i].marrital.marryflay)
{
case 1:
printf("\t 未婚\n");
break;
case 2:
printf("\t 已婚");
printf("\t%4d/%2d/%2d", woker[i].marrital.married.marrday.year, woker[i].marrital.married.marrday.month, woker[i].marrital.married.marrday.day);
printf("%8s %2d\n", woker[i].marrital.married.spousename, woker[i].marrital.married.children);
break;
case 3:
printf("\t 离婚");
printf("\t%4d/%2d/%2d", woker[i].marrital.divorce.divorcedata.year, woker[i].marrital.divorce.divorcedata.month, woker[i].marrital.divorce.divorcedata.day);
printf("\t\t%d\n", woker[i].marrital.divorce.children);
break;
}
}
}
