What is ultimately a time_t typedef to?

Question

I searched in linux box and saw it being typedef to

typedef __time_t time_t;

But could not find the __time_t definition.

Answer 1

It's a 32-bit signed integer type on most legacy platforms. However, that causes your code to suffer from the year 2038 bug. So modern C libraries should be defining it to be a signed 64-bit int instead, which is safe for a few billion years.

Answer 2

[root]# cat time.c

#include <time.h>

int main(int argc, char** argv)
{
        time_t test;
        return 0;
}

[root]# gcc -E time.c | grep __time_t

typedef long int __time_t;

It's defined in $INCDIR/bits/types.h through:

# 131 "/usr/include/bits/types.h" 3 4
# 1 "/usr/include/bits/typesizes.h" 1 3 4
# 132 "/usr/include/bits/types.h" 2 3 4

Answer 3

The time_t Wikipedia article article sheds some light on this. The bottom line is that the type of time_t is not guaranteed in the C specification.

The time_t datatype is a data type in the ISO C library defined for storing system time values. Such values are returned from the standard time() library function. This type is a typedef defined in the standard header. ISO C defines time_t as an arithmetic type, but does not specify any particular type, range, resolution, or encoding for it. Also unspecified are the meanings of arithmetic operations applied to time values.

Unix and POSIX-compliant systems implement the time_t type as a signed integer (typically 32 or 64 bits wide) which represents the number of seconds since the start of the Unix epoch: midnight UTC of January 1, 1970 (not counting leap seconds). Some systems correctly handle negative time values, while others do not. Systems using a 32-bit time_t type are susceptible to the Year 2038 problem.

Answer 4

Under Visual Studio 2008, it defaults to an __int64 unless you define _USE_32BIT_TIME_T. You're better off just pretending that you don't know what it's defined as, since it can (and will) change from platform to platform.