Re: Type conversion

manik <manik.budhiraja@xxxxxxxxx> writes:

Assuming int is 32 bits and char is 8 bits on my machine

Say I have the following piece of code

int a = 0x12345678;
char b;

b=(char)a;

b = a;

is simpler and has the same effect. Casts are for cases where type
conversion do not otherwise take place.

What would be the value of b?

Even the value of a in doubt! Let's assume that the initialisation of
a does not overflow...

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

No, not in any usual sense of the word.[1]

If char is unsigned the value is guaranteed by the standard (it is the
value of a mod UCHAR_MAX + 1) but the unsigned-ness of char is not
guaranteed, so the code is not portable. If char is signed, and the
value of a does not fit (it might!) then the result is implementation
defined.

[1] I say this because (since the result may be implementation-defined)
the implementation is permitted to cause the value of b to depend on
the (settable) endian-ness of the machine, but that is not what one
usually means by the phrase.

--
Ben.
.

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