Question

I've got myself in a terrible situation her. Can anyone refactor this C code for me?

for (i = 0; i < year1; ++i) {
!(i % 4) ? (!(i % 100) ? (!(i % 400) ? (days1 += 366) : (days1 += 365)) :
(days1 += 366)) :
(days1 += 365);

Answer 1

it's supposed to compute leap years

Answer 2

it should go like

if i is divisible by 4 then
if i is divisible by 100 then
if i is divisible by 400 then
days1 += 366
else
days1 += 365
else
days1 += 366
else
days1 += 365

Answer 3

better use i % 4 == 0 instead of !(i % 4) so I won't get confused :-P

Answer 4

why won't you just code it that way?

Answer 5

I got too crazy with the ternary operator :-P

Answer 6

otherwise it will take 10 lines of if-else stuff

Answer 7

for (i = 0; i < year1; ++i) {
if (i%400 = 0) {
days1+= 366
} else if (i%100 = 0) {
days1 += 365
} else if (i%
......
......

Answer 8

I always hated reading code that was trying to save lines of text and cram too much into one fancy line. Compilers will take care of optimizing all that stuff. Just write simple readable code.

Answer 9

this is the whole routine:

void days_between_dates(void) {
int month1, day1, year1, month2, day2, year2;
int days1, days2;
int i;
days1 = 0;
days2 = 0;
printf("Enter a date (dd/mm/yyyy): ");
scanf("%d/%d/%d", &day1, &month1, &year1);
printf("Enter another date (dd/mm/yyyy): ");
scanf("%d/%d/%d", &day2, &month2, &year2);
for (i = 0; i < year1; ++i) {
if (i % 4 == 0) {
if (i % 100 == 0) {
if (i % 400 == 0) {
days1 += 366;
} else {
days1 += 365;
}
} else {
days1 += 366;
}
} else {
days1 += 365;
}
}
for (i = 0; i < year2; ++i) {
if (i % 4 == 0) {
if (i % 100 == 0) {
if (i % 400 == 0) {
days2 += 366;
} else {
days2 += 365;
}
} else {
days2 += 366;
}
} else {
days2 += 365;
}
}
for (i = 1; i <= month1; ++i) {
switch(i) {
case 1:
case 3:
case 5:
case 7:
case 8:
case 10:
case 12:
days1 += 31;
break;
case 2:
!(i % 4) ? (days1 += 29) : (days1 += 30);
default:
days1 += 30;
break;
}
}
for (i = 1; i <= month2; ++i) {
switch(i) {
case 1:
case 3:
case 5:
case 7:
case 8:
case 10:
case 12:
days2 += 31;
break;
case 2:
!(i % 4) ? (days2 += 29) : (days2 += 30);
default:
days2 += 30;
break;
}
}
days1 += day1;
days2 += day2;
printf("\nThere is/are %d day(s) between those dates.\n", abs(days2 - days1));
}

Answer 10

how do I make it smaller?

Answer 11

Put this code in a function:
for (i = 0; i < year1; ++i) { if (i % 4 == 0) { if (i % 100 == 0) { if (i % 400 == 0) { days1 += 366; } else { days1 += 365; } } else { days1 += 366; } } else { days1 += 365; } }

Answer 12

I wish there was some easy way to do this whole thing, like call some C standard library time function that returns dates, find their difference, and then return that difference etc.

Answer 13

Your whole switch statement seems like an overkill. You just have one exception case (where it is 2nd month). Just write:
if (i=2) {
do this stuff;
} else {
do this other stuff
}

Answer 14

I've got two different cases: a month has 30 days or 31 days, and one exception case for february

Answer 15

I'm just glad the whole thing works :-P

Answer 16

Apparently, difftime() routine that is part of the ctime library returns number of seconds between two dates. Take a look.

Answer 17

Sorry, I now see your switch statement logic. Again, place that into a function instead of repeating same code twice.

Answer 18

If only C had lambda expressions :(