64 bit

[Guest]

C++ has no specify how many bytes is "int"

Until now, in most of 32 bit processors, "int" has 4 bytes
is it possible with new 64 bit processors (& compilers) "int" takes 8 bytes?

thanks

[John Harrison]

Re: 64 bit


"<- Chameleon ->" <cham_gss@hotma il.NOSPAM.com> wrote in message
news:bgagrm$3ri $1@nic.grnet.gr ...[color=blue]
> C++ has no specify how many bytes is "int"
>
> Until now, in most of 32 bit processors, "int" has 4 bytes
> is it possible with new 64 bit processors (& compilers) "int" takes 8[/color]
bytes?[color=blue]
>
> thanks
>[/color]

Yes it is.

There's no completely adequate solution to this problem in C++. I believe
this is addressed by the C99 standard, ask on news:comp.lang. c if you are
interested in this. Presumably one day the C99 standards will be applied to
C++ as well.

john

[Kevin Goodsell]

Re: 64 bit

<- Chameleon -> wrote:
[color=blue]
> C++ has no specify how many bytes is "int"
>
> Until now, in most of 32 bit processors, "int" has 4 bytes
> is it possible with new 64 bit processors (& compilers) "int" takes 8 bytes?
>
> thanks
>
>[/color]

Basically the rules about the size of int are: It must be at least 16
bits long, able to represent values from -32767 to 32767 (which implies
the "at least 16 bits" rule), and there's a suggestion that it reflect
the "natural word size of the target machine" or something like that.
Also, it may not be longer than long or shorter than short.

Note that int can be as little as 1 byte (without violating the rules
above), meaning sizeof(int) == 1. "bytes" in C++ do not need to be 8
bits long. A byte may be 32 bits long, for example (and I believe there
are implementations where this is the case). On such an implementation
you may have sizeof(char) == sizeof(short) == sizeof(int) ==
sizeof(long) == 1.

-Kevin