How to get rid of the warning with time.h in C++?

Question

When I use this

#include<time.h>
//...
int n = time(0);
//...

I get a warning about converting time to int. Is there a way to remove this warning?

Answer 1

Time returns time_t and not integer. Use that type preferably because it may be larger than int.

If you really need int, then typecast it explicitly, for example:

int n = (int)time(0);

Answer 2

Yes, change n to be a time_t. If you look at the signature in time.h on most / all systems, you'll see that that's what it returns.

#include<time.h>
//...
time_t n = time(0);
//...

Note that Arak is right: using a 32 bit int is a problem, at a minimum, due to the 2038 bug. However, you should consider that any sort of arithmetic on an integer n (rather than a time_t) only increases the probability that your code will trip over that bug early.

PS: In case I didn't make it clear in the original answer, the best response to a compiler warning is almost always to address the situation that you're being warned about. For example, forcing higher precision data into a lower precision variable loses information - the compiler is trying to warn you that you might have just created a landmine bug that someone won't trip over until much later.

Answer 3

I think you are using Visual C++. The return type of time(0) is 64bit int even if you are programming for 32bit platform unlike g++. To remove the warning, just assign time(0) to 64bit variable.

Answer 4

You probably want to use a type of time_t instead of an int.

See the example at http://en.wikipedia.org/wiki/Time_t.

Answer 5

The reason is time() functions returns a time_t time so you might need to do a static cast to an int or uint in this case. Write in this way:

time_t timer;
int n = static_cast<int> (time(&timer)); // this will give you current time as an integer and it is same as time(NULL)