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Re: gcc unsigned vs signed compare
On Wed, 31 Aug 2011, Manuel Bouyer wrote:
In lib/libc/gen/fixunsgen_ieee754.c there is a compare:
(exp - mant_dig > sizeof(UINTXX_T)*8-1)
exp is int, mant_dig is size_t (unsigned long) and sizeof
returns size_t. gcc generates an unsigned compare for this,
which doesn't gives the expected result when exp < mant_dig. Is
gcc right when generating an unsigned compare here ?
I think gcc is correct. C99 section 126.96.36.199 "Usual arithmetic
conversions" says: "Otherwise, if the operand that has unsigned
integer type has rank greater or equal to the rank of the type
of the other operand, then the operand with signed integer type
is converted to the type of the operand with unsigned integer
type." Here, mant_dig is the operand of the "-" opetator that has
unsigned type, and exp is the operand with signed type. "Rank"
is defined in section 188.8.131.52, and the rank of "unsigned long" is
greater than the rank of "int". So, the int operand is converted
to unsigned long before the subtraction, and the result of the
subtraction is unsigned long. sizeof returns an unsigned result,
so both operands of the ">" operator are unsigned.
--apb (Alan Barrett)
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