Quick C++ question

[Alphathree33]

Does

sint16 base[12];

declare an array of TWELVE 16-bit signed integers?

(For a total of 12x16 bits)?



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[MegaVovaN]

yes 12 integers numbered from 0 to 11

all will have default value (0)

edit: I didn't get to sint16 part, but I know arrays.

sint16 shouldn't be any different that int or double or string in this case

 

[dighn]

Originally posted by: MegaVovaN
yes 12 integers numbered from 0 to 11

all will have default value (0)

edit: I didn't get to sint16 part, but I know arrays.

sint16 shouldn't be any different that int or double or string in this case

not necessarily true. primitive values on stack are not initialized.

to the OP: if sint16 is defined to be a signed 16-bit integer then yeah. but it's non-standard.

and... you do know we have a forum for this stuff right?

 

[DaveSimmons]

At least for Visual C++ you can't count on variables to be initialized to any non-garbage value by the runtime, except class objects will run the constructor.

That will get you 12 "sint16"s, whatever sint16 is typedef'd to be, [0] - [11], filled with who-knows-what until you initialize them.

 

[MegaVovaN]

Sorry, Dave is right = they won't be 0 unless you write it like this:


sint16 base[12] = {0};

 

[Descartes]

Everyone else has pretty much covered it, but the ANSI C (99) and ANSI C++ (whatever revision it's at) all refer to the state of uninitialized variables as "indeterminate." Most platforms will throw exceptions when accessing unitialized space. *nix throws seg faults, and Windows platforms typically threw memory access violations.

Anyway, MegaVovaN's way of initializing an array is exactly how I've always done it.

 

[sao123]

we spent years getting a programming forum... please use it...
Lest they take it away again.