有没有办法在一行C代码中查找给定日期的星期几？
例如
鉴于19-05-2011（dd-mm-yyyy）给了我星期四
答案 0 :(得分：23)
正如Wikipedia报道的那样，1990年迈克尔基思和汤姆克拉弗发表了一个表达式，以尽量减少进入一个自包含函数所需的击键次数，以便将格里高利日期转换为一周中的数字日。
该表达式既不保留y也不保留d，并返回表示日期的从零开始的索引，从星期日开始，即如果该日是星期一，则表达式返回1。
使用表达式的代码示例如下：
int d = 15 ; //Day 1-31
int m = 5 ; //Month 1-12`
int y = 2013 ; //Year 2013`
int weekday = (d += m < 3 ? y-- : y - 2, 23*m/9 + d + 4 + y/4- y/100 + y/400)%7;
表达式使用逗号运算符，如this answer中所述。
享受！ ; - ）
答案 1 :(得分：16)
不太可能使用单行，但可以使用strptime function来解析您的日期格式，并且可以在系统上查询struct tm成员的tm_wday参数自动修改这些字段（例如一些glibc实现）。
int get_weekday(char * str) {
struct tm tm;
memset((void *) &tm, 0, sizeof(tm));
if (strptime(str, "%d-%m-%Y", &tm) != NULL) {
time_t t = mktime(&tm);
if (t >= 0) {
return localtime(&t)->tm_wday; // Sunday=0, Monday=1, etc.
}
}
return -1;
}
或者您可以对这些规则进行编码，以便在很长的单行中进行一些算术运算：
编辑：请注意，此解决方案仅适用于UNIX纪元（1970-01-01T00：00：00Z）之后的日期。
答案 2 :(得分：6)
这是基于wikipedia's article about Julian Day
的C99版本#include <stdio.h>
const char *wd(int year, int month, int day) {
/* using C99 compound literals in a single line: notice the splicing */
return ((const char *[]) \
{"Monday", "Tuesday", "Wednesday", \
"Thursday", "Friday", "Saturday", "Sunday"})[ \
( \
day \
+ ((153 * (month + 12 * ((14 - month) / 12) - 3) + 2) / 5) \
+ (365 * (year + 4800 - ((14 - month) / 12))) \
+ ((year + 4800 - ((14 - month) / 12)) / 4) \
- ((year + 4800 - ((14 - month) / 12)) / 100) \
+ ((year + 4800 - ((14 - month) / 12)) / 400) \
- 32045 \
) % 7];
}
int main(void) {
printf("%d-%02d-%02d: %s\n", 2011, 5, 19, wd(2011, 5, 19));
printf("%d-%02d-%02d: %s\n", 2038, 1, 19, wd(2038, 1, 19));
return 0;
}
通过从wd（）函数中的return行中删除拼接和空格，可以将其压缩为286个字符的单行：）
答案 3 :(得分：1)
这是我的实施。它很短，包括错误检查。如果您想要在01-01-1900之前的日期，您可以轻松地将锚更改为公历的开始日期。
#include <stdio.h>
int main(int argv, char** arv) {
int month[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
char* day[] = {"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"};
int d, m, y, i;
printf("Fill in a date after 01-01-1900 as dd-mm-yyyy: ");
scanf("%d-%d-%d", &d, &m, &y);
// correction for leap year
if (y % 4 == 0 && (y % 100 != 0 || y % 400 == 0))
month[1] = 29;
if (y < 1900 || m < 1 || m > 12 || d < 1 || d > month[m - 1]) {
printf("This is an invalid date.\n");
return 1;
}
for (i = 1900; i < y; i++)
if (i % 4 == 0 && (i % 100 != 0 || i % 400 == 0))
d += 366;
else
d += 365;
for (i = 0; i < m - 1; i++)
d += month[i];
printf("This is a %s.\n", day[d % 7]);
return 0;
}
答案 4 :(得分：1)
/*
Program to calculate the day on a given date by User
*/
#include<stdio.h>
#include<conio.h>
#include<process.h>
void main()
{
int dd=0,mm=0,i=0,yy=0,odd1=0,todd=0;//variable declaration for inputing the date
int remyr=0,remyr1=0,lyrs=0,oyrs=0,cyr=0,upyr=0,leap=0;//variable declaration for calculation of odd days
int montharr[12]={31,28,31,30,31,30,31,31,30,31,30,31};//array of month days
clrscr();
printf("Enter the date as DD-MM-YY for which you want to know the day\t:");
scanf("%d%d%d",&dd,&mm,&yy); //input the date
/*
check out correct date or not?
*/
if(yy%100==0)
{
if(yy%400==0)
{
//its the leap year
leap=1;
if(dd>29&&mm==2)
{
printf("You have entered wrong date");
getch();
exit(0);
}
}
else if(dd>28&&mm==2)
{
//not the leap year
printf("You have entered wrong date");
getch();
exit(0);
}
}
else if(yy%4==0)
{
//again leap year
leap=1;
if(dd>29&mm==2)
{
printf("You have entered wrong date");
getch();
exit(0);
}
}
else if(dd>28&&mm==2)
{
//not the leap year
printf("You have entered wrong date");
getch();
exit(0);
}
//if the leap year feb month contains 29 days
if(leap==1)
{
montharr[1]=29;
}
//check date,month,year should not be beyond the limits
if((mm>12)||(dd>31)|| (yy>5000))
{
printf("Your date is wrong");
getch();
exit(0);
}
//odd months should not contain more than 31 days
if((dd>31 && (mm == 1||mm==3||mm==5||mm==7||mm==8||mm==10||mm==12)))
{
printf("Your date is wrong");
getch();
exit(0);
}
//even months should not contains more than 30 days
if((dd>30 && (mm == 4||mm==6||mm==9||mm==11)))
{
printf("Your date is wrong");
getch();
exit(0);
}
//logic to calculate odd days.....
printf("\nYou have entered date: %d-%d-%d ",dd,mm,yy);
remyr1=yy-1;
remyr=remyr1%400;
cyr=remyr/100;
if(remyr==0)
{
oyrs=0;
}
else if(cyr==0 && remyr>0)
{
oyrs=0;
}
else if(cyr==1)
{
oyrs=5;
}
else if(cyr==2)
{
oyrs=3;
}
else if(cyr==3)
{
oyrs=1;
}
upyr=remyr%100;
lyrs=upyr/4;
odd1=lyrs+upyr;
odd1=odd1%7;
odd1=odd1+oyrs;
for(i=0;i<mm-1;i++)
{
odd1=odd1+montharr[i];
}
todd=odd1+dd;
if(todd>7)
todd=todd%7; //total odd days gives the re quired day....
printf("\n\nThe day on %d-%d-%d :",dd,mm,yy);
if(todd==0)
printf("Sunday");
if(todd==1)
printf("Monday");
if(todd==2)
printf("Tuesday");
if(todd==3)
printf("Wednesday");
if(todd==4)
printf("Thrusday");
if(todd==5)
printf("Friday");
if(todd==6)
printf("Saturday");
getch();
}
答案 5 :(得分：1)
我想出的答案：
const int16_t TM_MON_DAYS_ACCU[12] = {
0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334
};
int tm_is_leap_year(unsigned year) {
return ((year & 3) == 0) && ((year % 400 == 0) || (year % 100 != 0));
}
// The "Doomsday" the the day of the week of March 0th,
// i.e the last day of February.
// In common years January 3rd has the same day of the week,
// and on leap years it's January 4th.
int tm_doomsday(int year) {
int result;
result = TM_WDAY_TUE;
result += year; // I optimized the calculation a bit:
result += year >>= 2; // result += year / 4
result -= year /= 25; // result += year / 100
result += year >>= 2; // result += year / 400
return result;
}
void tm_get_wyday(int year, int mon, int mday, int *wday, int *yday) {
int is_leap_year = tm_is_leap_year(year);
// How many days passed since Jan 1st?
*yday = TM_MON_DAYS_ACCU[mon] + mday + (mon <= TM_MON_FEB ? 0 : is_leap_year) - 1;
// Which day of the week was Jan 1st of the given year?
int jan1 = tm_doomsday(year) - 2 - is_leap_year;
// Now just add these two values.
*wday = (jan1 + *yday) % 7;
}
使用这些定义（匹配struct tm的{{1}}）：
time.h答案 6 :(得分：1)
#include<stdio.h>
#include<math.h>
#include<conio.h>
int fm(int date, int month, int year) {
int fmonth, leap;
if ((year % 100 == 0) && (year % 400 != 0))
leap = 0;
else if (year % 4 == 0)
leap = 1;
else
leap = 0;
fmonth = 3 + (2 - leap) * ((month + 2) / (2 * month))+ (5 * month + month / 9) / 2;
fmonth = fmonth % 7;
return fmonth;
}
int day_of_week(int date, int month, int year) {
int dayOfWeek;
int YY = year % 100;
int century = year / 100;
printf("\nDate: %d/%d/%d \n", date, month, year);
dayOfWeek = 1.25 * YY + fm(date, month, year) + date - 2 * (century % 4);
//remainder on division by 7
dayOfWeek = dayOfWeek % 7;
switch (dayOfWeek) {
case 0:
printf("weekday = Saturday");
break;
case 1:
printf("weekday = Sunday");
break;
case 2:
printf("weekday = Monday");
break;
case 3:
printf("weekday = Tuesday");
break;
case 4:
printf("weekday = Wednesday");
break;
case 5:
printf("weekday = Thursday");
break;
case 6:
printf("weekday = Friday");
break;
default:
printf("Incorrect data");
}
return 0;
}
int main() {
int date, month, year;
printf("\nEnter the year ");
scanf("%d", &year);
printf("\nEnter the month ");
scanf("%d", &month);
printf("\nEnter the date ");
scanf("%d", &date);
day_of_week(date, month, year);
return 0;
}
输出： 输入2012年
输入02月
输入日期29
日期：2012年2月29日
工作日=星期三
答案 7 :(得分：0)
我认为你可以在glib中找到它： http://developer.gnome.org/glib/unstable/glib-Date-and-Time-Functions.html#g-date-get-day
此致
答案 8 :(得分：0)
这个有效：我把2006年1月作为参考。 （这是一个星期天）
int isLeapYear(int year) {
if(((year%4==0)&&(year%100!=0))||((year%400==0)))
return 1;
else
return 0;
}
int isDateValid(int dd,int mm,int yyyy) {
int isValid=-1;
if(mm<0||mm>12) {
isValid=-1;
}
else {
if((mm==1)||(mm==3)||(mm==5)||(mm==7)||(mm==8)||(mm==10)||(mm==12)) {
if((dd>0)&&(dd<=31))
isValid=1;
} else if((mm==4)||(mm==6)||(mm==9)||(mm==11)) {
if((dd>0)&&(dd<=30))
isValid=1;
} else {
if(isLeapYear(yyyy)){
if((dd>0)&&dd<30)
isValid=1;
} else {
if((dd>0)&&dd<29)
isValid=1;
}
}
}
return isValid;
}
int calculateDayOfWeek(int dd,int mm,int yyyy) {
if(isDateValid(dd,mm,yyyy)==-1) {
return -1;
}
int days=0;
int i;
for(i=yyyy-1;i>=2006;i--) {
days+=(365+isLeapYear(i));
}
printf("days after years is %d\n",days);
for(i=mm-1;i>0;i--) {
if((i==1)||(i==3)||(i==5)||(i==7)||(i==8)||(i==10)) {
days+=31;
}
else if((i==4)||(i==6)||(i==9)||(i==11)) {
days+=30;
} else {
days+= (28+isLeapYear(i));
}
}
printf("days after months is %d\n",days);
days+=dd;
printf("days after days is %d\n",days);
return ((days-1)%7);
}
答案 9 :(得分：0)
#include<stdio.h>
static char day_tab[2][13] = {
{0,31,28,31,30,31,30,31,31,30,31,30,31},
{0,31,29,31,30,31,30,31,31,30,31,30,31}
};
int main()
{
int year,month;
scanf("%d%d%d",&year,&month,&day);
printf("%d\n",day_of_year(year,month,day));
return 0;
}
int day_of_year(int year ,int month,int day)
{
int i,leap;
leap = year%4 == 0 && year%100 != 0 || year%400 == 0;
if(month < 1 || month >12)
return -1;
if (day <1 || day > day_tab[leap][month])
return -1;
for(i= 1;i<month ; i++)
{
day += day_tab[leap][year];
}
return day;
}
答案 10 :(得分：0)
int dayOfWeek(int y, int m, int d) {
return ((y-=m<3)+y/4-y/100+y/400+"-bed=pen+mad."[m]+d)%7;
}
来源 - here ↗
详情 - here ↗
答案 11 :(得分：0)
这是我在c中创建的一个简单代码，可以解决您的问题：
#include <conio.h>
int main()
{
int y,n,oy,ly,td,a,month,mon_,d,days,down,up; // oy==ordinary year, td=total days, d=date
printf("Enter the year,month,date: ");
scanf("%d%d%d",&y,&month,&d);
n= y-1; //here we subtracted one year because we have to find on a particular day of that year, so we will not count whole year.
oy= n%4;
if(oy==0) // for leap year
{
mon_= month-1;
down= mon_/2; //down means months containing 30 days.
up= mon_-down; // up means months containing 31 days.
if(mon_>=2)
{
days=(up*31)+((down-1)*30)+29+d; // here in down case one month will be of feb so we subtracted 1 and after that seperately
td= (oy*365)+(ly*366)+days; // added 29 days as it is the if block of leap year case.
}
if(mon_==1)
{
days=(up*31)+d;
td= (oy*365)+(ly*366)+days;
}
if(mon_==0)
{
days= d;
td= (oy*365)+(ly*366)+days;
}
}
else
{
mon_= month-1;
down= mon_/2;
up= mon_-down;
if(mon_>=2)
{
days=(up*31)+((down-1)*30)+28+d;
td= (oy*365)+(ly*366)+days;
}
if(mon_==1)
{
days=(up*31)+d;
td= (oy*365)+(ly*366)+days;
}
if(mon_==0)
{
days= d;
td= (oy*365)+(ly*366)+days;
}
}
ly= n/4;
a= td%7;
if(a==0)
printf("\nSunday");
if(a==1)
printf("\nMonday");
if(a==2)
printf("\nTuesday");
if(a==3)
printf("\nWednesday");
if(a==4)
printf("\nThursday");
if(a==5)
printf("\nFriday");
if(a==6)
printf("\nSaturday");
return 0;
}
答案 12 :(得分：0)
对于星期几，2000年至2099年。
uint8_t rtc_DayOfWeek(uint8_t year, uint8_t month, uint8_t day)
{
//static const uint8_t month_offset_table[] = {0, 3, 3, 6, 1, 4, 6, 2, 5, 0, 3, 5}; // Typical table.
// Added 1 to Jan, Feb. Subtracted 1 from each instead of adding 6 in calc below.
static const uint8_t month_offset_table[] = {0, 3, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4};
// Year is 0 - 99, representing years 2000 - 2099
// Month starts at 0.
// Day starts at 1.
// Subtract 1 in calc for Jan, Feb, only in leap years.
// Subtracting 1 from year has the effect of subtracting 2 in leap years, subtracting 1 otherwise.
// Adding 1 for Jan, Feb in Month Table so calc ends up subtracting 1 for Jan, Feb, only in leap years.
// All of this complication to avoid the check if it is a leap year.
if (month < 2) {
year--;
}
// Century constant is 6. Subtract 1 from Month Table, so difference is 7.
// Sunday (0), Monday (1) ...
return (day + month_offset_table[month] + year + (year >> 2)) % 7;
} /* end rtc_DayOfWeek() */
答案 13 :(得分：-1)
不在一行代码中，没有什么可以处理C标准库中的日期。但是，基于Doomsday算法或类似算法编写函数会相当简单。
答案 14 :(得分：-2)
#include<stdio.h>
int main(void) {
int n,y;
int ly=0;
int mon;
printf("enter the date\n");
scanf("%d",&n);
printf("enter the month in integer\n");
scanf("%d",&mon);
mon=mon-1;
printf("enter year\n");
scanf("%d",&y);
int dayT;
dayT=n%7;
if((y%4==0&&y%100!=0)|(y%4==0&&y%100==0&&y%400==0))
{
ly=y;
printf("the given year is a leap year\n");
}
char a[12]={6,2,2,5,0,3,5,1,4,6,2,4};
if(ly!=0)
{
a[0]=5;
a[1]=1;
}
int m,p;
m=a[mon];
int i,j=0,t=1;
for(i=1600;i<=3000;i++)
{
i=i+99;
if(i<y)
{
if(t==1)
{
p=5;t++;
}
else if(t==2)
{
p=3;
t++;
}
else if(t==3)
{
p=1;
t++;
}
else
{
p=0;
t=1;
}
}}
int q,r,s;
q=y%100;
r=q%7;
s=q/4;
int yTerm;
yTerm=p+r+s;
int w=dayT+m+yTerm;
w=w%7;
if(w==0)
printf("SUNDAY");
else if(w==1)
printf("MONDAY");
else if(w==2)
printf("TUESDAY");
else if(w==3)
printf("WEDNESDAY");
else if(w==4)
printf("THURSDAY");
else if(w==5)
printf("FRIDAY");
else
printf("SATURDAY");
return 0;
}