Format specifier to print time(0) in C

Question

We can declare a variable to hold the current time from the system using:

time_t now = time(0);

time(0) can also be use in genera

Answer 1

As already said, standard does not specifies what time_t is. If you want to be sure to use it in a portable way, first convert it to long and then print it as a long. That way your program won't suffer the Year 2038 problem even if 32 bits architectures are still available at that time :-) ...

time_t t = time(NULL);
long lt = (t > 0) ? t : (unsigned int) t;
printf("%ld", lt);

It should work for time_t being a 32 bits signed of insigned integer or a 64 bit integer. My formule will break when time_t will be a negative 64 bits integer, but contact me at that time and I should then be wise enough to find a better solution :-)

EDIT :

As noted by chux this is bad if int is only 16 bits. Here is a robust even if ugly way

union {
    long l;
    unsigned char[1] c;
} ul;
time_t t = time(NULL);
ul.l = 0;
ul.c[0] = 255;
if (ul.l < 0) { /* endianness test */
    ul.l = 0;
    memcpy(&ul.l, t&, sizeof(time_t));
}
else {
    ul.l = 0;
    memcpy(&ul.l + sizeof(long) - sizeof(time_t), &t, sizeof(time_t));
}
printf("%lu", (unsigned long) ul.l);

It should work in any configuration provided sizeof(long)>=sizeof(time_t)

EDIT 2:

chux's proposal seems to work on any reasonable cases. There are certainly pathological cases where it could break but on a normal implementation

time_t t = time(NULL);
long lt = (t > 0) ? t : t * 1u;
printf("%ld", lt);

should give correct result.