How to correctly cast time_t to long int?

Question

I'm still learning about type casting in C++ and I'm currently doing this

long int t = time(NULL);

I'm using VS2013 and noticed the conversion from 'time_t' to 'long' warning so I thought I would type cast it to look like;

long int t = static_cast<long int> time(NULL);

However this doesn't work yet combining a static cast and a C-style cast works

long int t = static_cast<long int> (time(NULL));

I was just wondering if anyone could help shed some light on this?

Answer 1

time(NULL) is not a cast but a function call which returns time_t. Since time_t is not exactly the same type as long int, you see the warning.

Furthermore, static_cast<T>(value) requires the parenthesis, that is why your first version does not work.

Answer 2

Your question contains the answer. The static_cast generic method in the code you provide takes the time_t type as its input and converts it to a long int as its return value. This code does not contain a C-style type-cast.

long int t = static_cast<long int> (time(NULL));

Type-casting should also work too, because time_t is an arithmetic type and the C cast operator will perform the promotion to the long int type.

long int t = (long int)time(NULL);

This casting tutorial might be an interesting read for you.