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    SubjectRe: [patch -next] x86, microcode, AMD: signedness bug in generic_load_microcode()
    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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