8 replies to this topic

[nuclear123]

how would i do this?

sizeof(variable) == sizeof(char[5])

is there a replacement i could use for sizeof(char[5]) which would signify 5 bytes?
for example....a char is not always == to a byte...even tho on most compiler's it is....this can change in the future..my question is how can i check the a certain variable has 8 bits being used and is by definition a byte

[nobodynews]

http://www.cplusplus.com/reference/clibrary/climits/

CHAR_BIT seems like the it would work. You'd want this, I think:

sizeof(variable) * CHAR_BIT == 5*8

[nuclear123]

is there anyways c++ actually allows messing with actual BIT's to do comparison on sizes? i feel this would be a more secure/safe/portable way -thx

[karwosts]

is there anyways c++ actually allows messing with actual BIT's to do comparison on sizes? i feel this would be a more secure/safe/portable way -thx

Nope, don't think so. Bytes are the unit of memory. You can't get an address of a bit.

[SiS-Shadowman]

From the wiki entry to stdint I gather that you can rely on the uintN_t types to be integers that hold exactly N bytes. Thus sizeof(5 * uint8_t) should do the job. However you need to verify if the platforms you require support at least this part of the C99 standard.

[magic_man]

how would i do this?

sizeof(variable) == sizeof(char[5])

is there a replacement i could use for sizeof(char[5]) which would signify 5 bytes?

Yes the literal 5

for example....a char is not always == to a byte...even tho on most compiler's it is....this can change in the future..my question is how can i check the a certain variable has 8 bits being used and is by definition a byte

No a char is always a Language byte, how many bits this has is a different matter and one which CHAR_BIT will tell you.

[Hodgman]

is there a replacement i could use for sizeof(char[5]) which would signify 5 bytes?

Yes the literal 5

Yep, first thing wikipedia says is that sizeof is a number of bytes, so you want sizeof(variable) == 5

In the programming languages C and C++, the unary operator 'sizeof' is used to calculate the sizes of datatypes, in number of bytes

[samoth]

Why would one care anyway? Unless you write something very, very, very specific, you should not need to worry. First, it is not likely for you to ever encounter a non-8 bit system, and even so... if bytes are not 8 bits, they still are "the unit", a char is still a byte, and the size of any other POD or struct is still a multiple of char.

So... don't make your life complicated if you don't have to.

[Zahlman]

how would i do this?

sizeof(variable) == sizeof(char[5])

is there a replacement i could use for sizeof(char[5]) which would signify 5 bytes?

Yes. It's called "5".

for example....a char is not always == to a byte...

Yes, it is. However, a byte is not always the same thing as an octet.

If a char has 64 bits in your environment, then a byte has 64 bits in your environment. As far as the language of the C++ standard is concerned, there is nothing about a byte that involves having exactly 8 bits in it.

even tho on most compiler's it is....this can change in the future..

It has to do with more than just your compiler.

my question is how can i check the a certain variable has 8 bits being used

A char must have at least 8 bits. It is also not allowed to have a type smaller than char. Every variable's size is made up of char-sized units. (sizeof(char) == 1, by definition).

The real question is, why do you care about the exact number of bits in use by any of your data types? Even if you knew and had a good reason to care, there is very little you can do about it at this high of a level of programming. Yes, for the exceedingly few people who have a reason to care about this sort of thing, C++ (or even C) is probably much too high-level.