在 2018/9/20 23:05, Vincent Lefevre 写道:
> Actually, there may be a bug with -m32:
>
> #include <stdio.h>
> #include <float.h>
>
> int main (void)
> {
> float f = 2147483646;
> printf ("FLT_EVAL_METHOD = %d\n", (int) FLT_EVAL_METHOD);
> if (f == 2147483647)
> {
> printf("%.20g\n", (double) f);
> }
> return 0;
> }
>
> $ gcc -std=c99 -m32 -O tst.c -o tst
> $ ./tst
> FLT_EVAL_METHOD = 2
> 2147483648
>
> Since FLT_EVAL_METHOD = 2, float_t is long double, thus shouldn't
> 2147483647 be converted to long double directly (as f is regarded
> as an operand of type long double), so that the result of the
> equality test should be false?
>
> The C standard says for FLT_EVAL_METHOD = 2: "evaluate all operations
> and constants to the range and precision of the long double type".
> ^^^^^^^^^^^^^
>
-----
ISO/IEC 9899:2017
(WG14 N2176)
5.2.4.2.2 Characteristics of floating types <float.h>
9 Except for assignment and cast (which remove all extra range and
precision), the values yielded by operators with floating operands and
values subject to the usual arithmetic conversions and of **floating
constants** are evaluated to a format whose range and precision may be
greater than required by the type. The use of evaluation formats is
characterized by the implementation-defined value of FLT_EVAL_METHOD:24)
-----
`2147483647` is an integer constant. This rule only describes floating
constants, so it does not apply.
According to '6.3.1.8 Usual arithmetic conversions', here `2147483647`
is converted to a value having type `float`, which is then compared with
`f` using the internal `long double` type.
> Converting f first to float is contrary to the idea behind
> FLT_EVAL_METHOD = 2, which is to avoid loss of precision in
> intermediate operations.
>
--
Best regards,
LH_Mouse
