atan2f gives different results with m32 flag - c

I'm porting some code from 32 bit to 64 bit, and ensuring the answers are the same. In doing so, I noticed that atan2f was giving different results between the two.
I created this min repro:
#include <stdio.h>
#include <math.h>
void testAtan2fIssue(float A, float B)
{
float atan2fResult = atan2f(A, B);
printf("atan2f: %.15f\n", atan2fResult);
float atan2Result = atan2(A, B);
printf("atan2: %.15f\n", atan2Result);
}
int main()
{
float A = 16.323556900024414;
float B = -5.843180656433105;
testAtan2fIssue(A, B);
}
When built with:
gcc compilerTest.c -m32 -o 32bit.out -lm
it gives:
atan2f: 1.914544820785522
atan2: 1.914544820785522
When built with:
gcc compilerTest.c -o 64bit.out -lm
it gives:
atan2f: 1.914544701576233
atan2: 1.914544820785522
Note that atan2 gives the same result in both cases, but atan2f does not.
Things I have tried:
Building the 32 bit version with -ffloat-store
Building the 32 bit version with -msse2 -mfpmath=sse
Building the 64 bit version with -mfpmath=387
None changed the results for me.
(All of these were based on the hypothesis that it has something to do with the way floating point operations happen on 32 bit vs 64 bit architectures.)
Question:
What are my options for getting them to give the same result? (Is there a compiler flag I could use?) And also, what is happening here?
I'm running on an i7 machine, if that is helpful.

This is easier to see in hex notation.
void testAtan2fIssue(float A, float B) {
double d = atan2(A, B);
printf(" atan2 : %.13a %.15f\n", d, d);
float f = atan2f(A, B);
printf(" atan2f: %.13a %.15f\n", f, f);
printf("(float) atan2 : %.13a %.15f\n", (float) d, (float) d);
float f2 = nextafterf(f, 0);
printf("problem value : %.13a %.15f\n", f2, f2);
}
// _ added for clarity
atan2 : 0x1.ea1f9_b9d85de4p+0 1.914544_797857041
atan2f: 0x1.ea1f9_c0000000p+0 1.914544_820785522
(float) atan2 : 0x1.ea1f9_c0000000p+0 1.914544_820785522
problem value : 0x1.ea1f9_a0000000p+0 1.914544_701576233
what is happening here?
The conversion from double to float can be expected to be optimal, yet arctangent functions may be a few ULP off on various platforms. The 1.914544701576233 is the next smaller float value and reflects the slightly inferior arctangent calculation.
What are my options for getting them to give the same result?
Few. Code could roll your own my_atan2() from an established code base. Yet even that may have subtle implementation differences. #stark
Instead, consider making code checking tolerant of the minute variations.

Related

Why doesn't the same generated assembler code lead to the same output?

Sample code (t0.c):
#include <stdio.h>
float f(float a, float b, float c) __attribute__((noinline));
float f(float a, float b, float c)
{
return a * c + b * c;
}
int main(void)
{
void* p = V;
printf("%a\n", f(4476.0f, 20439.0f, 4915.0f));
return 0;
}
Invocation & execution (via godbolt.org):
# icc 2021.1.2 on Linux on x86-64
$ icc t0.c -fp-model=fast -O3 -DV=f
0x1.d32322p+26
$ icc t0.c -fp-model=fast -O3 -DV=0
0x1.d32324p+26
Generated assembler code is the same: https://godbolt.org/z/osra5jfYY.
Why doesn't the same generated assembler code lead to the same output?
Why does void* p = f; matter?
Godbolt shows you the assembly emitted by running the compiler with -S. But in this case, that's not the code that actually gets run, because further optimizations can be done at link time.
Try checking the "Compile to binary" box instead (https://godbolt.org/z/ETznv9qP4), which will actually compile and link the binary and then disassemble it. We see that in your -DV=f version, the code for f is:
addss xmm0,xmm1
mulss xmm0,xmm2
ret
just as before. But with -DV=0, we have:
movss xmm0,DWORD PTR [rip+0x2d88]
ret
So f has been converted to a function which simply returns a constant loaded from memory. At link time, the compiler was able to see that f was only ever called with a particular set of constant arguments, and so it could perform interprocedural constant propagation and have f merely return the precomputed result.
Having an additional reference to f evidently defeats this. Probably the compiler or linker sees that f had its address taken, and didn't notice that nothing was ever done with the address. So it assumes that f might be called elsewhere in the program, and therefore it has to emit code that would give the correct result for arbitrary arguments.
As to why the results are different: The precomputation is done strictly, evaluating both a*c and b*c as float and then adding them. So its result of 122457232 is the "right" one by the rules of C, and it is also what you get when compiling with -O0 or -fp-model=strict. The runtime version has been optimized to (a+b)*c, which is actually more accurate because it avoids an extra rounding; it yields 122457224, which is closer to the exact value of 122457225.

Same FLT_EVAL_METHOD, different results in GCC/Clang

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.

C program to find roots error

I am writing a function in C with the below specifications:
float find_root(float a, float b, float c, float p, float q);
find_root takes the coefficients a,b,c of a quadratic equation and an interval (p, q). It will return the root of this equation in the given interval.
For example: find_root(1, -8, 15, 2, 4) should produce a root "close to" 3.0
I have written the below code, and I don't understand why it doesn't work:
#include<stdio.h>
#include<math.h>
main()
{
printf("Hello World");
}
float find_root(float a, float b, float c, float p, float q) {
float d,root1,root2;
d = b * b - 4 * a * c;
root1 = ( -b + sqrt(d)) / (2* a);
root2 = ( -b - sqrt(d)) / (2* a);
if (root1<=q || root1>=p)
{
return root1;
}
return root2;
}
Please let me know what the error is.
Your program doesn't work, because, you never called find_root() from your main().
find_root() is not suppossed to run all-by-itself. Your program statrs execution from main(). You need to call your sub-function from main() in order to make them execute.
Change your main to have a call to find_root(), something like below.
int main() //put proper signature
{
float anser = 0;
answer = find_root(1, -8, 15, 2, 4); //taken from the question
printf("The anser is %f\n", answer); //end with a \n, stdout is line buffered
return 0; //return some value, good practice
}
Then, compile the program like
gcc -o output yourfilename.c -lm
Apart from this, for the logical issue(s) in find_root() function, please follow the way suggested by Mr. #paxdiablo.
For that data, your two roots are 5 and 3. With p == 2 and q == 4:
if (root1<=q || root1>=p)
becomes:
if (5<=4 || 5>=2)
which is true, so you'll get 5.
The if condition you want is:
if ((p <= root1) && (root1 <= q))
as shown in the following program, that produces the correct 3:
#include<stdio.h>
#include<math.h>
float find_root (float a, float b, float c, float p, float q) {
float d,root1,root2;
d = b * b - 4 * a * c;
root1 = ( -b + sqrt(d)) / (2* a);
root2 = ( -b - sqrt(d)) / (2* a);
if ((p <= root1) && (root1 <= q))
return root1;
return root2;
}
int main (void) {
printf ("%f\n", find_root(1, -8, 15, 2, 4));
return 0;
}
That's the logic errors with your calculations of the roots.
Just keep in mind there are other issues with your code.
You need to ensure you actually call the function itself, your main as it stands does not.
It also wont produce a value within the p/q bounds, instead it will give you the first root if it's within those bounds, otherwise it'll give you the second root regardless of its value.
You may want to catch the situation where d is negative, since you don't want to take the square root of it:
a = 1000, b = 0, c = 1000: d <- -4,000,000
And, lastly, if your compiler is complaining about not being able to link sqrt (as per one of your comments), you'll probably find you can fix that by specifying the math library, something like:
gcc -o myprog myprog.c -lm
Your program starts at main by definition.
Your main function is not calling find_root but it should.
You need to compile with all warnings & debug info (gcc -Wall -Wextra -g) then use a debugger (gdb) to run your code step by step to understand the behavior of your program, so compile with
gcc -Wall -Wextra -g yoursource.c -lm -o yourbinary
or with
clang -Wall -Wextra -g yoursource.c -lm -o yourbinary
then learn how to use gdb (e.g. run gdb ./yourbinary ... and later ./yourbinary without a debugger)
Then you'll think and improve the source code and recompile it, and debug it again. And repeat that process till you are happy with your program.
BTW, you'll better end your printf format strings with \n or learn about fflush(3)
Don't forget to read the documentation of every function (like printf(3) ...) that you are calling.
You might want to give some arguments (thru your main(int argc, char**argv) ...) to your program. You could use atof(3) to convert them to a double
Read also about undefined behavior, which you should always avoid.
BTW, you can use any standard C compiler (and editor like emacs or gedit) for your homework, e.g. use gcc or clang on your Linux laptop (then use gdb ...). You don't need a specific seashell
Change this condition
if (root1<=q || root1>=p)
to
if (root1<=q && root1>=p)
otherwise if anyone of the conditions is satisfied, root1 will be returned and root2 will almost never be returned. Hope this fixes your problem.

Is there a document describing how Clang handles excess floating-point precision?

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.

double-precision numbers in inline assembly (GCC, IA-32)

I'm just starting to learn assembly and I want to round a floating-point value using a specified rounding mode. I've tried to implement this using fstcw, fldcw, and frndint.
Right now I get a couple of errors:
~ $ gc a02p
gcc -Wall -g a02p.c -o a02p
a02p.c: In function `roundD':
a02p.c:33: error: parse error before '[' token
a02p.c:21: warning: unused variable `mode'
~ $
I'm not sure if I am even doing this right at all. I don't want to use any predefined functions. I want to use GCC inline assembly.
This is the code:
#include <stdio.h>
#include <stdlib.h>
#define PRECISION 3
#define RND_CTL_BIT_SHIFT 10
// floating point rounding modes: IA-32 Manual, Vol. 1, p. 4-20
typedef enum {
ROUND_NEAREST_EVEN = 0 << RND_CTL_BIT_SHIFT,
ROUND_MINUS_INF = 1 << RND_CTL_BIT_SHIFT,
ROUND_PLUS_INF = 2 << RND_CTL_BIT_SHIFT,
ROUND_TOWARD_ZERO = 3 << RND_CTL_BIT_SHIFT
} RoundingMode;
double roundD (double n, RoundingMode roundingMode)
{
short c;
short mode = (( c & 0xf3ff) | (roundingMode));
asm("fldcw %[nIn] \n"
"fstcw %%eax \n" // not sure why i would need to store the CW
"fldcw %[modeIn] \n"
"frndint \n"
"fistp %[nOut] \n"
: [nOut] "=m" (n)
: [nIn] "m" (n)
: [modeIn] "m" (mode)
);
return n;
}
int main (int argc, char **argv)
{
double n = 0.0;
if (argc > 1)
n = atof(argv[1]);
printf("roundD even %.*f = %.*f\n",
PRECISION, n, PRECISION, roundD(n, ROUND_NEAREST_EVEN));
printf("roundD down %.*f = %.*f\n",
PRECISION, n, PRECISION, roundD(n, ROUND_MINUS_INF));
printf("roundD up %.*f = %.*f\n",
PRECISION, n, PRECISION, roundD(n, ROUND_PLUS_INF));
printf("roundD zero %.*f = %.*f\n",
PRECISION, n, PRECISION, roundD(n, ROUND_TOWARD_ZERO));
return 0;
}
Am I even remotely close to getting this right?
A better process is to write a simple function that rounds a floating point value. Next, instruct your compiler to print an assembly listing for the function. You may want to put the function in a separate file.
This process will show you the calling and exiting conventions used by the compiler. By placing the function in a separate file, you won't have to build other files. Also, it will give you the opportunity to replace the C language function with an assembly language function.
Although inline assembly is supported, I prefer to replace an entire function in assembly language and not use inline assembly (inline assembly isn't portable, so the source will have to be changed when porting to a different platform).
GCC's inline assembler syntax is arcane to say the least, and I do not claim to be an expert, but when I have used it I used this howto guide. In all examples all template markers are of the form %n where n is a number, rather then the %[ttt] form that you have used.
I also note that the line numbers reported in your error messages do not seem to correspond with the code you posted. So I wonder if they are in fact for this exact code?

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