c - 如何将三个无符号整数打包成 C 中的一个无符号短整数

Question

我有一个大学作业，其中我必须使用 C 打包三个代表日期的无符号整数(年份必须在 1970 到 2097 之间)并将其压缩成一个无符号短整数，然后再次解压并在命令行中显示。

有人可以帮帮我吗？

下面我留下我目前的代码(函数外声明的变量和函数签名不能更改)

编辑:我有第二个问题，即 getchar 函数不返回任何内容，将三个日期字段设置为 0...

#include <stdio.h>

typedef unsigned short pkdate_t;
typedef struct date3 {
    unsigned year;
    unsigned month;
    unsigned day;
} date3_t;

int ror (int val, unsigned n) {
    return (val >> n)|(n << (32 - n));
}

int pack_date (pkdate_t * dst, date3_t * src) {
    if ((*src).year < 1097 || (*src).year > 2097 || (*src).month < 1 || (*src).month > 12 || (*src).day < 1 ||
    ((*src).month == 2 && (((*src).year / 4 == 0 && (*src).day > 29) || ((*src).year / 4 != 0 && (*src).day > 28))) ||
    (((*src).month == 2 || (*src).month == 4 || (*src).month == 6 || (*src).month == 9 || (*src).month == 11) && (*src).day == 31)) return -1;

    //(*dst) = (*src).year * 10 + (*src).month + (*src).day; "wrong code"

    return 0;
}

int unpack_date (date3_t * dst, pkdate_t date) {
    (*dst).year = date % 10000;
    date = date / 10000;
    (*dst).month = date % 100;
    date = date / 100;
    (*dst).day = date % 100;

    if ((*dst).year < 1097 || (*dst).year > 2097 || (*dst).month < 1 || (*dst).month > 12 || (*dst).day < 1 ||
    ((*dst).month == 2 && (((*dst).year / 4 == 0 && (*dst).day > 29) || ((*dst).year / 4 != 0 && (*dst).day > 28))) ||
    (((*dst).month == 2 || (*dst).month == 4 || (*dst).month == 6 || (*dst).month == 9 || (*dst).month == 11) && (*dst).day == 31)) return -1;
    else return 0;
}

int main () {
    int k = ror(30, 2);
    printf("%s\n", "exercicio 1:");
    printf("%d turned into %d\n", 30, k);

    pkdate_t date = 0;
    date3_t newDate;
    newDate.year = 2000;
    newDate.month = 04;
    newDate.day = 02;
    printf("%s\n", "exercicio 2:");
    //printf("%s\n", "insira uma data (yyyymmdd) e de seguida pressione enter:");

    //int i = 1000;

    //while (i > 0) {
        //newDate.year += getchar() * i;
        //i = i / 10;
    //}

    //i = 10;

    //while (i > 0) {
        //newDate.month += getchar() * i;
        //i = i / 10;
    //}

    //i = 10;

    //while (i > 0) {
        //newDate.day += getchar() * i;
        //i = i / 10;
    //}

    //"the commented part above does not get any values from command line, still figuring that part out :)"

    pack_date(&date, &newDate);

    printf("packed date: %hu\n", date);

    unpack_date(&newDate, date);

    printf("unpacked date: %u/%u/%u\n", newDate.year, newDate.month, newDate.day);

    //newDate.year = 2000;
    //newDate.month = 12;
    //newDate.day = 14;
    //pack_date(&date, &newDate);
    //printf("%s\n", "exercicio 3:");
    //printf("date is %hu\n", date);

    // "ignore"

    return 0;
}

Answer 1

在 C 语言中，没有那么多符合位域条件的用例，但是恕我直言，这个非常适合，因为它可以让您免去很多位移位带来的麻烦:

struct date_struct
{
    unsigned short year:7;
    unsigned short month:4;
    unsigned short day:5;
};

union date
{
   struct date_struct bits;
   short date;
}