about time and time_t

Thanks GNiewerth,

In fact time_t is a typedef of either int or long, I dont remember right now. You can safely convert time_t to (unsigned) long and vice versa.

Your answer is helpful.

regards,
George

Thanks treuss,

There is a reason why there are separate types like time_t and size_t, so I would always recommend not casting to maintain portability. size_t is equivalent to unsigned long on most of todays systems, both 32bit and 64bit, but there is no guarantee it is and will be like that on all system. time_t will have to be 64bit for any system that wants to survive the next 30 years, but not every system will have to implement a 32bit long type and I am sure there will be some embedded systems which will not.

Why do you want to cast it anyways, why not use time_t whenever you have a time_t?

Because I mixed time_t with other numeric data types (in some numeric formula), like long, so I have to do the conversion. Any comments?

regards,
George

Because I mixed time_t with other numeric data types (in some numeric formula), like long, so I have to do the conversion. Any comments?I wouldn't say so. You can add any integer to a time_t and the result will again be time_t. If you take the difference between two time_t's the result might be an integer of which only the application knows how big it has to be and if signed or unsigned. So the following does calculations but no conversions:time_t today = time(0);
time_t yesterday = today - 86400;
unsigned int distance_of_all_my_trouble = today - yesterday;

Thanks treuss,

I wouldn't say so. You can add any integer to a time_t and the result will again be time_t. If you take the difference between two time_t's the result might be an integer of which only the application knows how big it has to be and if signed or unsigned. So the following does calculations but no conversions:time_t today = time(0);
time_t yesterday = today - 86400;
unsigned int distance_of_all_my_trouble = today - yesterday;

Sorry that in my post, conversion means compiler warnings, for example, your code has the following warnings in Visual Studio 2005,

warning C4244: 'initializing' : conversion from 'time_t' to 'unsigned int', possible loss of data

regards,
George

Thanks Viggy,

Adding a cast you mean,

unsigned_int_type_variable = (unsigned int) (time_t_type_variable);
~~~~~~~~~

Right (see the *~~~~* part)? ;)

Right, because an 'unsigned int' is not strictly a 'time_t'. If you add a cast, then the warning goes away because you are telling the compiler that you know that these are different types.

Viggy

regards,
George

Thanks GNiewerth,

You should use the C++ static_cast:

unsigned int uiTime = static_cast<unsigned int>( time( NULL ) );

What is the benefit of your static cast method campared with my method below?

unsigned_int_type_variable = (unsigned int) (time_t_type_variable);

regards,
George

Hi GNiewerth,

#1 you can easily spot there is a cast because of the explicit static_cast keyword.

#2 its safer than the C-style cast. C-style cast just convert anything into anything else, where C++ style casts have some requirements that prevent you from performing invalid casts.

I am programming C. ;)

But I am very interested to learn #2. Could you show me a sample about how static_cast prevents invalid type conversion please? Or recommend me some URLs?

regards,
George

Thanks GNiewerth,

CodeProject article on static_cast ( http://www.codeproject.com/cpp/static_cast.asp)

Type casting article on cplusplus.com ( http://www.cplusplus.com/doc/tutorial/typecasting.html)

Very useful information.

regards,
George