Format specifiers for implementation-defined types like time_t

Question

Usually you can use a cast to convert the operand to some type for which you know the right format.

Your proposed solution:

time_t t = time(NULL);
printf("%s", t);

clearly will not work, since time_t is a numeric type, not char*.

We know, in general, that time_t is an arithmetic type. Something like this:

 printf("%ld\n", (long)t);

is likely to work on most systems. It can fail (a) if time_t is an unsigned type no wider than unsigned long and the current value of t exceeds LONG_MAX, or (b) if time_t is a floating-point type.

If you have C99 support, you can use long long, which is a little better:

printf("%lld\n", (long long)t);

If you really want to go overboard with portability, you can detect what kind of type time_t is:

if ((time_t)-1 > 0) {
    // time_t is an unsigned type
    printf("%ju\n", (uintmax_t)t);
}
else if ((time_t)1 / 2 > 0) {
    // time_t is a signed integer type
    printf("%jd\n", (intmax_t)t);
}
else {
    // time_t is a floating-point type (I've never seen this)
    printf("%Lg\n", (long double)t);
}

You might want to tweak the %Lg format to something like %Lf or %.10Lf, depending on what output format you want.

Again, this assumes C99 support -- and you'll need #include <stdint.h> to make uintmax_t and intmax_t visible.

time_t and clock_t are a bit unusual, in that the standard says only that they're arithmetic type capable of representing times. (In principle they could be complex types, but I'd say ignoring that possibility is worth the risk.)

In most other cases, you'll probably know whether a given type is signed, unsigned, or floating-point, and you can just convert to the widest type of that kind.

Note that if you don't know how time_t is represented, you probably won't understand the output of the printf (such as 1379375215) either -- unless your goal is to figure that out.

(If you were programming in C++ rather than C, std::cout << t << "\n"; would automatically use the correct overloaded operator<<.)

If you want human-readable output (like Mon 2013-09-16 16:46:55 PDT), you'll want to use one of the conversion functions declared in <time.h>, such as asctime() or strftime().

Generally, the way to display the value of a time_t is to break down its components to a struct tm using gmtime or localtime and display those or convert them as desired with strftime, or ctime to go directly from time_t to a string showing local time.

If you want to see the raw value for some purpose, the C standard specifies that time_t is real, which means it is integer or floating-point (C 2011 (N1570) 6.2.5 17). Therefore, you should be able to convert it to double and print that. There is some possibility that time_t can represent values that double cannot, so you might have to guard against that if you want to take care regarding exotic implementations. Since difftime returns the difference of two time_t objects as a double, it seems C does not truly support time_t with more precision than a double.

You can use difftime() to obtain a double:

time_t t = time(NULL);
printf("seconds 1970->now: %.f\n", difftime(t, (time_t) 0));

It is simple and I think it is portable.

The C standard says time_t will be a 'real type' (meaning an integer type or a floating point type, though in practice it is always an integer type).

With time_t, your best bet is to format it with strftime() after analyzing it with localtime() or gmtime() — this can be done portably.

Unportably, you have to determine by some mechanism what is the correct format specifier. You might use PRI_[Xxodi]_time and SCN_[Xxodi]_time or something similar as a non-standard but close-to-standard (without trampling on the reserved namespace — which is names starting PRI or SCN followed by a lower-case letter or X). You use some mechanism to specify that...encapsulating the unportable information in one place.