Re: Type conversion

In article <m3y7a6cfzp.fsf@xxxxxxxxxxxxxxx>,
Mark L Pappin <mlp@xxxxxxx> wrote:

int a = 0x12345678;
char b;
b=(char)a;

Would the value of b depend on the endianness of the machine?

Yes.

Casts, like most other operations, are defined in terms of value, not
representation, no? Thus the low-order 8 bits will be used, and in
this instance since
(a & 0xff) == (a & 0x7f)
even the signedness of 'char' is irrelevant as values up to 127 are
guaranteed representable even in 'signed char'.

If char is signed, we have a conversion to a signed integer with
smaller size, and in that case if the value cannot be represented
the result is implementation-defined.

-- Richard
--
:wq
.

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