Why do I get the same value from sin(x) and sinf(x) as in (8) and (9) below? Why do I get different values of [ x - sinf(x) ] from two different implementations, (2) and (3) below? Why (6) gives the same result as (5)?
I am using g++ (Ubuntu/Linaro 4.6.3-1ubuntu5) 4.6.3. I used -O0 to disable optimization.
Background: I was tracking down a bug in my program where I am required to use float throughout the program because it will be ported to an embedded system. However, I am currently debugging on Ubuntu on my PC just because it is convenient. I found the operation like (xs) caused inaccuracy when x was small. That made me think it must be due to loss of significant digits by catastrophic cancellation. However, when I replaced a variable, s, with sinf(x), the problem of inaccuracy did not occur (as seen in (2) vs (3)). I could guess sinf() might be implemented as the same thing as sin(). If so, why does the explicit casting to float have no effect as in (4) and (5). Now I am puzzled.
int main()
{
unsigned long int xx(0x3d65c2f2);
float x(*reinterpret_cast<float*>(&xx));
float s(sinf(x));
printf("( 1) x = %.10e\n", x);
printf("( 2) x - s = %.10e\n", x-s);
printf("( 3) x - sinf(x) = %.10e\n", x-sinf(x)); // Why is it different from (2)?
printf("( 4) x - float(sinf(x)) = %.10e\n", x-float(sinf(x))); // Compare with (3). Why casting has no effect?
printf("( 5) float(x) - float(sinf(x)) = %.10e\n", float(x)-float(sinf(x))); // Compare with (3). Why casting has no effect?
printf("( 6) x - sin(x) = %.10e\n", x - sin(x));
printf("( 7) s = %.10e\n", s);
printf("( 8) sinf(x) = %.10e\n", sinf(x));
printf("( 9) sin(x) = %.10e\n", sin(x)); // Compare with (8). Is sinf() identical to sin()?
printf("(10) float(sinf(x)) = %.10e\n", float(sinf(x))); // Compare with (8). Why casting has no effect?
double s_df(sinf(x));
double s_dd(sin(x));
float s_fd(sin(x));
float s_ff(sinf(x));
printf("(20) s_df = %.10e\n", s_df);
printf("(21) s_dd = %.10e\n", s_dd); // Compare with (20). Is sinf() identical to sin()?
printf("(22) s_fd = %.10e\n", s_fd);
printf("(23) s_ff = %.10e\n", s_ff);
return 0;
}
Here is the output:
$ make && ./main
g++ main.cc -Wall -c -o main.o -O0
g++ -o main main.o
( 1) x = 5.6094117463e-02
( 2) x - s = 2.9411166906e-05
( 3) x - sinf(x) = 2.9412529899e-05
( 4) x - float(sinf(x)) = 2.9412529899e-05
( 5) float(x) - float(sinf(x)) = 2.9412529899e-05
( 6) x - sin(x) = 2.9412529899e-05
( 7) s = 5.6064706296e-02
( 8) sinf(x) = 5.6064704933e-02
( 9) sin(x) = 5.6064704933e-02
(10) float(sinf(x)) = 5.6064704933e-02
(20) s_df = 5.6064704933e-02
(21) s_dd = 5.6064704933e-02
(22) s_fd = 5.6064706296e-02
(23) s_ff = 5.6064706296e-02
In C++, sin
has an overload float sin(float f)
. And overload resolution is done on argument type, rather than return type. To force the use of double sin(double d)
you need to cast the argument: sin(static_cast<double>(x))
.
(2) vs (3): the FP standard allows implementations to store intermediate results with greater accuracy than the final result. So the value of s
need not be exactly the same as the intermediate result for sin(f)
in (3).
A lot of this is dependent on your compiler, compiler settings and hardware. For example, if I run your code on my system I get:
( 1) x = 5.6094117463e-02
( 2) x - s = 2.9411166906e-05
( 3) x - sinf(x) = 2.9411166906e-05
( 4) x - float(sinf(x)) = 2.9411166906e-05
( 5) float(x) - float(sinf(x)) = 2.9411166906e-05
( 6) x - sin(x) = 2.9412529899e-05
( 7) s = 5.6064706296e-02
( 8) sinf(x) = 5.6064706296e-02
( 9) sin(x) = 5.6064704933e-02
(10) float(sinf(x)) = 5.6064706296e-02
(20) s_df = 5.6064706296e-02
(21) s_dd = 5.6064704933e-02
(22) s_fd = 5.6064706296e-02
(23) s_ff = 5.6064706296e-02
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