Sample code (t50.c):
#include <stdio.h>
#include <stdint.h>
#include <inttypes.h>
#include <math.h>
#include <float.h>
#include <assert.h>
const float d1 = NAN;
const float d2 = -0x0p+0;
const float d3 = NAN / -0x0p+0;
typedef union { uint32_t u; float d; } u_t;
int main(void)
{
u_t u1;
u_t u2;
u_t u3;
u1.d = *(volatile float*)&d1 / *(volatile float*)&d2;
u2.d = d3;
u3.d = d1 / d2;
if ( u1.u != u2.u || u1.u != u3.u )
{
printf("error:\n");
printf("u1 (run time) %08"PRIx32" %.*e\n", u1.u, DECIMAL_DIG, u1.d);
printf("u2 (compile time) %08"PRIx32" %.*e\n", u2.u, DECIMAL_DIG, u2.d);
printf("u3 %08"PRIx32" %.*e\n", u3.u, DECIMAL_DIG, u3.d);
}
return 0;
}
Compiler invocation: cl t50.c /O1 /fp:precise && t50
Expected result: <nothing>
Actual result (same for cl x86 and cl x64):
error:
u1 (run time) ffc00000 -nan(ind)
u2 (compile time) 7f800000 inf
u3 ffc00000 -nan(ind)
I've specified /fp:strict: cl t50.c /O1 /fp:strict && t50, but got:
t50.c(8): error C2099: initializer is not a constant
t50.c(10): error C2099: initializer is not a constant
cl version: 19.25.28611 for x86 and 19.25.28611 for x64.
Compare with gcc (10.2.0) and clang (11.0.0):
gcc t50.c -O2 && a.exe
<nothing>
clang t50.c -O2 && a.exe
<nothing>
Why? What I'm missing here?
What the standard (C / IEEE) says?
UPD1:
Same result for if ( u1.d != u2.d || u1.d != u3.d ).
Same result w/o unions (i.e. using float u1, u2, u3).
UPD2:
compiler's NAN definition: Windows Kits\10\Include\10.0.18362.0\ucrt\corecrt_math.h: #define NAN ((float)(INFINITY * 0.0F)).
Output for u1.d = NAN; printf("NAN %08"PRIx32" %.*e\n", u1.u, DECIMAL_DIG, u1.d); (for both cl x86 and cl x64): NAN ffc00000 -nan(ind).
Why?
cl does not follow IEEE-754 concerning NAN.
What I'm missing here?
Assuming a compliant C compiler follows IEEE.
What the standard C says?
C is lax with specs in the NAN department. I'd rate the compiler as compliant, just a weak quality of implementation concerning NAN.
What the standard IEEE says?
Non-compliant.
My question today should not be very much complicated, but I simply can't find a reason/solution. As a small, reproducible example, consider the following toy C code
#define _state_ 0
#if _state_ == 1
int foo(void) {return 1;}
#else
/* check GCC flags first
note that -mfma will automatically turn on -mavx, as shown by [gcc -mfma -dM -E - < /dev/null | egrep "SSE|AVX|FMA"]
so it is sufficient to check for -mfma only */
#ifndef __FMA__
#error "Please turn on GCC flag: -mfma"
#endif
#include <immintrin.h> /* All OK, compile C code */
void foo (double *A, double *B) {
__m256d A1_vec = _mm256_load_pd(A);
__m256d B_vec = _mm256_broadcast_sd(B);
__m256d C1_vec = A1_vec * B_vec;
}
#endif
I am going to compile this test.c file by
gcc -fpic -O2 -c test.c
Note I did not turn on GCC flag -mfma, so the #error will be triggered. What I would expect, is that compilation will stop immediately after GCC sees this #error, but this is what I got with GCC 5.3:
test.c:14:2: error: #error "Please turn on GCC flag: -mfma"
#error "Please turn on GCC flag: -mfma"
^
test.c: In function ‘foo’:
test.c:22:11: warning: AVX vector return without AVX enabled changes the ABI [-Wpsabi]
__m256d A1_vec = _mm256_load_pd(A);
^
GCC does stops, but why does it also pick up a line after #error? Any explanation? Thanks.
For people who want to try, there are some hardware requirement. You need an x86-64 with AVX and FMA instruction sets.
I have a draft copy of the C ISO spec, and in §4/4 - it states
The implementation shall not successfully translate a preprocessing translation unit containing a #error preprocessing directive unless it is part of a group skipped by conditional inclusion.
Later on, in §6.10.5, where #error is formally defined, it says
A preprocessing directive of the form
# error pp-tokens opt new-line
causes the implementation to produce a diagnostic message that includes the specified
sequence of preprocessing tokens.
In other words, the spec only requires that any code that has an #error just needs to fail to compile and report an error message along the way, not that the compilation immediately needs to terminate as soon as #error is reached.
Given that it's considered good practice to always check the top-level errors reported by a compiler before later ones, I'd imagine a competent programmer who saw a string of errors beginning with an #error directive would likely know what was going on.
The following program (adapted from here) is giving inconsistent results when compiled with GCC (4.8.2) and Clang (3.5.1). In particular, the GCC result does not change even when FLT_EVAL_METHOD does.
#include <stdio.h>
#include <float.h>
int r1;
double ten = 10.0;
int main(int c, char **v) {
printf("FLT_EVAL_METHOD = %d\n", FLT_EVAL_METHOD);
r1 = 0.1 == (1.0 / ten);
printf("0.1 = %a, 1.0/ten = %a\n", 0.1, 1.0 / ten);
printf("r1=%d\n", r1);
}
Tests:
$ gcc -std=c99 t.c && ./a.out
FLT_EVAL_METHOD = 0
0.1 = 0x1.999999999999ap-4, 1.0/ten = 0x1.999999999999ap-4
r1=1
$ gcc -std=c99 -mpfmath=387 t.c && ./a.out
FLT_EVAL_METHOD = 2
0.1 = 0x0.0000000000001p-1022, 1.0/ten = 0x0p+0
r1=1
$ clang -std=c99 t.c && ./a.out
FLT_EVAL_METHOD = 0
0.1 = 0x1.999999999999ap-4, 1.0/ten = 0x1.999999999999ap-4
r1=1
$ clang -std=c99 -mfpmath=387 -mno-sse t.c && ./a.out
FLT_EVAL_METHOD = 2
0.1 = 0x0.07fff00000001p-1022, 1.0/ten = 0x0p+0
r1=0
Note that, according to this blog post, GCC 4.4.3 used to output 0 instead of 1 in the second test.
A possibly related question indicates that a bug has been corrected in GCC 4.6, which might explain why GCC's result is different.
I would like to confirm if any of these results would be incorrect, or if some subtle evaluation steps (e.g. a new preprocessor optimization) would justify the difference between these compilers.
This answer is about something that you should resolve before you go further, because it is going to make reasoning about what happens much harder otherwise:
Surely printing 0.1 = 0x0.07fff00000001p-1022 or 0.1 = 0x0.0000000000001p-1022 can only be a bug on your compilation platform caused by ABI mismatch when using -mfpmath=387. None of these values can be excused by excess precision.
You could try to include your own conversion-to-readable-format in the test file, so that that conversion is also compiled with -mfpmath=387. Or make a small stub in another file, not compiled with that option, with a minimalistic call convention:
In other file:
double d;
void print_double(void)
{
printf("%a", d);
}
In the file compiled with -mfpmath=387:
extern double d;
d = 0.1;
print_double();
Ignoring the printf problem which Pascal Cuoq addressed, I think GCC is correct here: according to the C99 standard, FLT_EVAL_METHOD == 2 should
evaluate all operations and constants to the range and precision of the long double type.
So, in this case, both 0.1 and 1.0 / ten are being evaluated to an extended precision approximation of 1/10.
I'm not sure what Clang is doing, though this question might provide some help.
I'm doing some experiments with the new _Generic keyword and stumbled upon a special case regarding multiple evaluations. See the following:
#include <stdio.h>
#define write_char(c) _Generic(c, char: putchar, const char: putchar)(c)
int main(void)
{
const char *s = "foo";
write_char(*s++);
write_char(*s++);
write_char(*s++);
putchar('\n');
}
This compiles fine and produces the expected result with GCC:
$ gcc -std=c11 -Wall plusplus.c -o plusplus
$ ./plusplus
foo
On the other hand, Clang outputs a big honking warning:
$ clang -std=c11 plusplus.c -o plusplus
plusplus.c:9:18: warning: multiple unsequenced modifications to 's'
[-Wunsequenced]
write_char(*s++);
^~
plusplus.c:3:32: note: expanded from macro 'write_char'
#define write_char(c) _Generic(c, char: putchar, const char: putchar)(c)
...
Yet the result is as expected:
$ ./plusplus
foo
I checked the draft of the standard, which says (at p. 97 of the PDF):
The controlling expression of a generic selection is not evaluated.
This seems to precisely address the problem of side-effects in macros (e.g., MIN and MAX).
Now, can I safely ignore Clang's warning, or am I wrong?
As I mentioned in comments, you posted the question about two weeks after the bug was fixed in Clangs trunk. See revision rL223266 (December 3 2014). The fix is included in Clang 3.6.
Now, can I safely ignore Clang's warning, or am I wrong?
We already know that you're right, so here is a way to ignore pragmas in Clang for the
future:
#pragma clang diagnostic push
#pragma clang diagnostic ignored "-Wunsequenced"
write_char(*s++);
#pragma clang diagnostic pop
To do not repeat it on every use of the macro, you could put _Pragma in its body:
#define write_char(c) \
_Pragma("clang diagnostic push") \
_Pragma("clang diagnostic ignored \"-Wunsequenced\"") \
_Generic(c, char: putchar, const char: putchar)(c) \
_Pragma("clang diagnostic pop")
It seems that was a bug. It has now been solved since clang 3.6 onward as shown here.
It is nearly impossible(*) to provide strict IEEE 754 semantics at reasonable cost when the only floating-point instructions one is allowed to used are the 387 ones. It is particularly hard when one wishes to keep the FPU working on the full 64-bit significand so that the long double type is available for extended precision. The usual “solution” is to do intermediate computations at the only available precision, and to convert to the lower precision at more or less well-defined occasions.
Recent versions of GCC handle excess precision in intermediate computations according to the interpretation laid out by Joseph S. Myers in a 2008 post to the GCC mailing list. This description makes a program compiled with gcc -std=c99 -mno-sse2 -mfpmath=387 completely predictable, to the last bit, as far as I understand. And if by chance it doesn't, it is a bug and it will be fixed: Joseph S. Myers' stated intention in his post is to make it predictable.
Is it documented how Clang handles excess precision (say when the option -mno-sse2 is used), and where?
(*) EDIT: this is an exaggeration. It is slightly annoying but not that difficult to emulate binary64 when one is allowed to configure the x87 FPU to use a 53-bit significand.
Following a comment by R.. below, here is the log of a short interaction of mine with the most recent version of Clang I have :
Hexa:~ $ clang -v
Apple clang version 4.1 (tags/Apple/clang-421.11.66) (based on LLVM 3.1svn)
Target: x86_64-apple-darwin12.4.0
Thread model: posix
Hexa:~ $ cat fem.c
#include <stdio.h>
#include <math.h>
#include <float.h>
#include <fenv.h>
double x;
double y = 2.0;
double z = 1.0;
int main(){
x = y + z;
printf("%d\n", (int) FLT_EVAL_METHOD);
}
Hexa:~ $ clang -std=c99 -mno-sse2 fem.c
Hexa:~ $ ./a.out
0
Hexa:~ $ clang -std=c99 -mno-sse2 -S fem.c
Hexa:~ $ cat fem.s
…
movl $0, %esi
fldl _y(%rip)
fldl _z(%rip)
faddp %st(1)
movq _x#GOTPCREL(%rip), %rax
fstpl (%rax)
…
This does not answer the originally posed question, but if you are a programmer working with similar issues, this answer might help you.
I really don't see where the perceived difficulty is. Providing strict IEEE-754 binary64 semantics while being limited to 80387 floating-point math, and retaining 80-bit long double computation, seems to follow well-specified C99 casting rules with both GCC-4.6.3 and clang-3.0 (based on LLVM 3.0).
Edited to add: Yet, Pascal Cuoq is correct: neither gcc-4.6.3 or clang-llvm-3.0 actually enforce those rules correctly for '387 floating-point math. Given the proper compiler options, the rules are correctly applied to expressions evaluated at compile time, but not for run-time expressions. There are workarounds, listed after the break below.
I do molecular dynamics simulation code, and am very familiar with the repeatability/predictability requirements and also with the desire to retain maximum precision available when possible, so I do claim I know what I am talking about here. This answer should show that the tools exist and are simple to use; the problems arise from not being aware of or not using those tools.
(A preferred example I like, is the Kahan summation algorithm. With C99 and proper casting (adding casts to e.g. Wikipedia example code), no tricks or extra temporary variables are needed at all. The implementation works regardless of compiler optimization level, including at -O3 and -Ofast.)
C99 explicitly states (in e.g. 5.4.2.2) that casting and assignment both remove all extra range and precision. This means that you can use long double arithmetic by defining your temporary variables used during computation as long double, also casting your input variables to that type; whenever a IEEE-754 binary64 is needed, just cast to a double.
On '387, the cast generates an assignment and a load on both the above compilers; this does correctly round the 80-bit value to IEEE-754 binary64. This cost is very reasonable in my opinion. The exact time taken depends on the architecture and surrounding code; usually it is and can be interleaved with other code to bring the cost down to neglible levels. When MMX, SSE or AVX are available, their registers are separate from the 80-bit 80387 registers, and the cast usually is done by moving the value to the MMX/SSE/AVX register.
(I prefer production code to use a specific floating-point type, say tempdouble or such, for temporary variables, so that it can be defined to either double or long double depending on architecture and speed/precision tradeoffs desired.)
In a nutshell:
Don't assume (expression) is of double precision just because all the variables and literal constants are. Write it as (double)(expression) if you want the result at double precision.
This applies to compound expressions, too, and may sometimes lead to unwieldy expressions with many levels of casts.
If you have expr1 and expr2 that you wish to compute at 80-bit precision, but also need the product of each rounded to 64-bit first, use
long double expr1;
long double expr2;
double product = (double)(expr1) * (double)(expr2);
Note, product is computed as a product of two 64-bit values; not computed at 80-bit precision, then rounded down. Calculating the product at 80-bit precision, then rounding down, would be
double other = expr1 * expr2;
or, adding descriptive casts that tell you exactly what is happening,
double other = (double)((long double)(expr1) * (long double)(expr2));
It should be obvious that product and other often differ.
The C99 casting rules are just another tool you must learn to wield, if you do work with mixed 32-bit/64-bit/80-bit/128-bit floating point values. Really, you encounter the exact same issues if you mix binary32 and binary64 floats (float and double on most architectures)!
Perhaps rewriting Pascal Cuoq's exploration code, to correctly apply casting rules, makes this clearer?
#include <stdio.h>
#define TEST(eq) printf("%-56s%s\n", "" # eq ":", (eq) ? "true" : "false")
int main(void)
{
double d = 1.0 / 10.0;
long double ld = 1.0L / 10.0L;
printf("sizeof (double) = %d\n", (int)sizeof (double));
printf("sizeof (long double) == %d\n", (int)sizeof (long double));
printf("\nExpect true:\n");
TEST(d == (double)(0.1));
TEST(ld == (long double)(0.1L));
TEST(d == (double)(1.0 / 10.0));
TEST(ld == (long double)(1.0L / 10.0L));
TEST(d == (double)(ld));
TEST((double)(1.0L/10.0L) == (double)(0.1));
TEST((long double)(1.0L/10.0L) == (long double)(0.1L));
printf("\nExpect false:\n");
TEST(d == ld);
TEST((long double)(d) == ld);
TEST(d == 0.1L);
TEST(ld == 0.1);
TEST(d == (long double)(1.0L / 10.0L));
TEST(ld == (double)(1.0L / 10.0));
return 0;
}
The output, with both GCC and clang, is
sizeof (double) = 8
sizeof (long double) == 12
Expect true:
d == (double)(0.1): true
ld == (long double)(0.1L): true
d == (double)(1.0 / 10.0): true
ld == (long double)(1.0L / 10.0L): true
d == (double)(ld): true
(double)(1.0L/10.0L) == (double)(0.1): true
(long double)(1.0L/10.0L) == (long double)(0.1L): true
Expect false:
d == ld: false
(long double)(d) == ld: false
d == 0.1L: false
ld == 0.1: false
d == (long double)(1.0L / 10.0L): false
ld == (double)(1.0L / 10.0): false
except that recent versions of GCC promote the right hand side of ld == 0.1 to long double first (i.e. to ld == 0.1L), yielding true, and that with SSE/AVX, long double is 128-bit.
For the pure '387 tests, I used
gcc -W -Wall -m32 -mfpmath=387 -mno-sse ... test.c -o test
clang -W -Wall -m32 -mfpmath=387 -mno-sse ... test.c -o test
with various optimization flag combinations as ..., including -fomit-frame-pointer, -O0, -O1, -O2, -O3, and -Os.
Using any other flags or C99 compilers should lead to the same results, except for long double size (and ld == 1.0 for current GCC versions). If you encounter any differences, I'd be very grateful to hear about them; I may need to warn my users of such compilers (compiler versions). Note that Microsoft does not support C99, so they are completely uninteresting to me.
Pascal Cuoq does bring up an interesting problem in the comment chain below, which I didn't immediately recognize.
When evaluating an expression, both GCC and clang with -mfpmath=387 specify that all expressions are evaluated using 80-bit precision. This leads to for example
7491907632491941888 = 0x1.9fe2693112e14p+62 = 110011111111000100110100100110001000100101110000101000000000000
5698883734965350400 = 0x1.3c5a02407b71cp+62 = 100111100010110100000001001000000011110110111000111000000000000
7491907632491941888 * 5698883734965350400 = 42695510550671093541385598890357555200 = 100000000111101101101100110001101000010100100001011110111111111111110011000111000001011101010101100011000000000000000000000000
yielding incorrect results, because that string of ones in the middle of the binary result is just at the difference between 53- and 64-bit mantissas (64 and 80-bit floating point numbers, respectively). So, while the expected result is
42695510550671088819251326462451515392 = 0x1.00f6d98d0a42fp+125 = 100000000111101101101100110001101000010100100001011110000000000000000000000000000000000000000000000000000000000000000000000000
the result obtained with just -std=c99 -m32 -mno-sse -mfpmath=387 is
42695510550671098263984292201741942784 = 0x1.00f6d98d0a43p+125 = 100000000111101101101100110001101000010100100001100000000000000000000000000000000000000000000000000000000000000000000000000000
In theory, you should be able to tell gcc and clang to enforce the correct C99 rounding rules by using options
-std=c99 -m32 -mno-sse -mfpmath=387 -ffloat-store -fexcess-precision=standard
However, this only affects expressions the compiler optimizes, and does not seem to fix the 387 handling at all. If you use e.g. clang -O1 -std=c99 -m32 -mno-sse -mfpmath=387 -ffloat-store -fexcess-precision=standard test.c -o test && ./test with test.c being Pascal Cuoq's example program, you will get the correct result per IEEE-754 rules -- but only because the compiler optimizes away the expression, not using the 387 at all.
Simply put, instead of computing
(double)d1 * (double)d2
both gcc and clang actually tell the '387 to compute
(double)((long double)d1 * (long double)d2)
This is indeed I believe this is a compiler bug affecting both gcc-4.6.3 and clang-llvm-3.0, and an easily reproduced one. (Pascal Cuoq points out that FLT_EVAL_METHOD=2 means operations on double-precision arguments is always done at extended precision, but I cannot see any sane reason -- aside from having to rewrite parts of libm on '387 -- to do that in C99 and considering IEEE-754 rules are achievable by the hardware! After all, the correct operation is easily achievable by the compiler, by modifying the '387 control word to match the precision of the expression. And, given the compiler options that should force this behaviour -- -std=c99 -ffloat-store -fexcess-precision=standard -- make no sense if FLT_EVAL_METHOD=2 behaviour is actually desired, there is no backwards compatibility issues, either.) It is important to note that given the proper compiler flags, expressions evaluated at compile time do get evaluated correctly, and that only expressions evaluated at run time get incorrect results.
The simplest workaround, and the portable one, is to use fesetround(FE_TOWARDZERO) (from fenv.h) to round all results towards zero.
In some cases, rounding towards zero may help with predictability and pathological cases. In particular, for intervals like x = [0,1), rounding towards zero means the upper limit is never reached through rounding; important if you evaluate e.g. piecewise splines.
For the other rounding modes, you need to control the 387 hardware directly.
You can use either __FPU_SETCW() from #include <fpu_control.h>, or open-code it. For example, precision.c:
#include <stdlib.h>
#include <stdio.h>
#include <limits.h>
#define FP387_NEAREST 0x0000
#define FP387_ZERO 0x0C00
#define FP387_UP 0x0800
#define FP387_DOWN 0x0400
#define FP387_SINGLE 0x0000
#define FP387_DOUBLE 0x0200
#define FP387_EXTENDED 0x0300
static inline void fp387(const unsigned short control)
{
unsigned short cw = (control & 0x0F00) | 0x007f;
__asm__ volatile ("fldcw %0" : : "m" (*&cw));
}
const char *bits(const double value)
{
const unsigned char *const data = (const unsigned char *)&value;
static char buffer[CHAR_BIT * sizeof value + 1];
char *p = buffer;
size_t i = CHAR_BIT * sizeof value;
while (i-->0)
*(p++) = '0' + !!(data[i / CHAR_BIT] & (1U << (i % CHAR_BIT)));
*p = '\0';
return (const char *)buffer;
}
int main(int argc, char *argv[])
{
double d1, d2;
char dummy;
if (argc != 3) {
fprintf(stderr, "\nUsage: %s 7491907632491941888 5698883734965350400\n\n", argv[0]);
return EXIT_FAILURE;
}
if (sscanf(argv[1], " %lf %c", &d1, &dummy) != 1) {
fprintf(stderr, "%s: Not a number.\n", argv[1]);
return EXIT_FAILURE;
}
if (sscanf(argv[2], " %lf %c", &d2, &dummy) != 1) {
fprintf(stderr, "%s: Not a number.\n", argv[2]);
return EXIT_FAILURE;
}
printf("%s:\td1 = %.0f\n\t %s in binary\n", argv[1], d1, bits(d1));
printf("%s:\td2 = %.0f\n\t %s in binary\n", argv[2], d2, bits(d2));
printf("\nDefaults:\n");
printf("Product = %.0f\n\t %s in binary\n", d1 * d2, bits(d1 * d2));
printf("\nExtended precision, rounding to nearest integer:\n");
fp387(FP387_EXTENDED | FP387_NEAREST);
printf("Product = %.0f\n\t %s in binary\n", d1 * d2, bits(d1 * d2));
printf("\nDouble precision, rounding to nearest integer:\n");
fp387(FP387_DOUBLE | FP387_NEAREST);
printf("Product = %.0f\n\t %s in binary\n", d1 * d2, bits(d1 * d2));
printf("\nExtended precision, rounding to zero:\n");
fp387(FP387_EXTENDED | FP387_ZERO);
printf("Product = %.0f\n\t %s in binary\n", d1 * d2, bits(d1 * d2));
printf("\nDouble precision, rounding to zero:\n");
fp387(FP387_DOUBLE | FP387_ZERO);
printf("Product = %.0f\n\t %s in binary\n", d1 * d2, bits(d1 * d2));
return 0;
}
Using clang-llvm-3.0 to compile and run, I get the correct results,
clang -std=c99 -m32 -mno-sse -mfpmath=387 -O3 -W -Wall precision.c -o precision
./precision 7491907632491941888 5698883734965350400
7491907632491941888: d1 = 7491907632491941888
0100001111011001111111100010011010010011000100010010111000010100 in binary
5698883734965350400: d2 = 5698883734965350400
0100001111010011110001011010000000100100000001111011011100011100 in binary
Defaults:
Product = 42695510550671098263984292201741942784
0100011111000000000011110110110110011000110100001010010000110000 in binary
Extended precision, rounding to nearest integer:
Product = 42695510550671098263984292201741942784
0100011111000000000011110110110110011000110100001010010000110000 in binary
Double precision, rounding to nearest integer:
Product = 42695510550671088819251326462451515392
0100011111000000000011110110110110011000110100001010010000101111 in binary
Extended precision, rounding to zero:
Product = 42695510550671088819251326462451515392
0100011111000000000011110110110110011000110100001010010000101111 in binary
Double precision, rounding to zero:
Product = 42695510550671088819251326462451515392
0100011111000000000011110110110110011000110100001010010000101111 in binary
In other words, you can work around the compiler issues by using fp387() to set the precision and rounding mode.
The downside is that some math libraries (libm.a, libm.so) may be written with the assumption that intermediate results are always computed at 80-bit precision. At least the GNU C library fpu_control.h on x86_64 has the comment "libm requires extended precision". Fortunately, you can take the '387 implementations from e.g. GNU C library, and implement them in a header file or write a known-to-work libm, if you need the math.h functionality; in fact, I think I might be able to help there.
For the record, below is what I found by experimentation. The following program shows various behaviors when compiled with Clang:
#include <stdio.h>
int r1, r2, r3, r4, r5, r6, r7;
double ten = 10.0;
int main(int c, char **v)
{
r1 = 0.1 == (1.0 / ten);
r2 = 0.1 == (1.0 / 10.0);
r3 = 0.1 == (double) (1.0 / ten);
r4 = 0.1 == (double) (1.0 / 10.0);
ten = 10.0;
r5 = 0.1 == (1.0 / ten);
r6 = 0.1 == (double) (1.0 / ten);
r7 = ((double) 0.1) == (1.0 / 10.0);
printf("r1=%d r2=%d r3=%d r4=%d r5=%d r6=%d r7=%d\n", r1, r2, r3, r4, r5, r6, r7);
}
The results vary with the optimization level:
$ clang -v
Apple LLVM version 4.2 (clang-425.0.24) (based on LLVM 3.2svn)
$ clang -mno-sse2 -std=c99 t.c && ./a.out
r1=0 r2=1 r3=0 r4=1 r5=1 r6=0 r7=1
$ clang -mno-sse2 -std=c99 -O2 t.c && ./a.out
r1=0 r2=1 r3=0 r4=1 r5=1 r6=1 r7=1
The cast (double) that differentiates r5 and r6 at -O2 has no effect at -O0 and for variables r3 and r4. The result r1 is different from r5 at all optimization levels, whereas r6 only differs from r3 at -O2.