Why does gettimeofday return weird tv_sec?

Question

I have an application that calls gettimeofday to store the tv_sec in which that part of the code starts to run. This code is very simple:

struct timeval tvnow;
gettimeofday(&tvnow);
int initialTime = tvnow.tv_sec;

It normally works fine but sometimes I am getting unexpected results such as

tvnow = {tv_sec = 1024, tv_usec = 0}
initialTime = 1401591

Or

tvnow = {tv_sec = 1024, tv_usec = 0}
initialTime = 2439903

Why could this happen?

Regards

What happens if you change that `int initialTime` to `time_t initialTime`? My guess is you're on a 64 bit computer and it is overflowing the int. — Adam D. Ruppe, May 12 '15 at 15:03
Post how you determined `initialTime` has the value of `1401591`. — chux - Reinstate Monica, May 12 '15 at 15:17

John Bollinger · Accepted Answer · 2015-05-12T15:24:07.383

You are calling gettimeofday() with the wrong number of arguments. The resulting behavior is undefined. If you don't want the timezone information, then pass NULL as the second argument:

struct timeval tvnow;
gettimeofday(&tvnow, NULL);
int initialTime = tvnow.tv_sec;

I'm going to suppose that somehow you're including a combination of headers that include a definition of struct timeval (or else you're providing your own definition), but not a prototype for gettimeofday(). If there were a correct prototype visible at the point of the function call, then the code oughtn't to compile at all. If your compiler doesn't issue a warning about the absence of a prototype then you ought to either figure out how to make it do so (and compile always that way), or get a better compiler.

Why does gettimeofday return weird tv_sec?

1 Answers1