On Sun, Feb 20, 2011 at 03:14:52PM +0100, Borislav Petkov wrote:
> int f() {
> return 0xa5a5a5a5;
> }
>
> int main()
> {
>
> char ret = f();
>
> printf("ret = 0x%016x\n", ret);
>
> return 0;
> }
> --
>
> doesn't cause a warning and prints a sign extended 0x00000000ffffffa5
> which is cast to the return type of the function. If ret is an unsigned
> char, then we return a 0x00000000000000a5.
>
> I found something about it in the C99 standard??, section "6.5.16.1 Simple
> assignment":
>
> 4. EXAMPLE 1 In the program fragment
>
> int f(void);
> char c;
> /* ... */
> if ((c = f()) == -1)
> /* ... */
>
> the int value returned by the function may be truncated when stored in
> the char, and then converted back to int width prior to the comparison.
> In an implementation in which ??????plain?????? char has the same range
> of values as unsigned char (and char is narrower than int), the result
> of the conversion cannot be negative, so the operands of the comparison
> can never compare equal. Therefore, for full portability, the variable c
> should be declared as int."
>
> so the whole "... may be truncated.. " could mean a lot of things. From
> my example above, gcc does truncate the int return type to a byte-sized
> char only when they differ in signedness.
No, that's not what's going on. GCC _is_ truncating to a byte, 0xa5,
whether it's signed or not. Then at the time of the call to printf,
the 0xa5 is cast to int. If the char is signed, 0xa5 is sign-extended;
if unsigned, it's zero-extended.
--
Matthew Wilcox Intel Open Source Technology Centre
"Bill, look, we understand that you're interested in selling us this
operating system, but compare it to ours. We can't possibly take such
a retrograde step."
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