plasmacel plasmacel - 2 months ago 21
C++ Question

Difference between ldexp(1, x) and exp2(x)

It seems if the floating-point representation has radix 2 (i.e.

FLT_RADIX == 2
) both
std::ldexp(1, x)
and
std::exp2(x)
raise
2
to the given power
x
.

Does the standard define or mention any expected behavioral and/or performance difference between them? What is the practical experience over different compilers?

Answer

exp2(x) and ldexp(x,i) perform two different operations. The former computes 2x, where x is a floating-point number, while the latter computes x*2i, where i is an integer. For integer values of x, exp2(x) and ldexp(1,int(x)) would be equivalent, provided the conversion of x to integer doesn't overflow.

The question about the relative efficiency of these two functions doesn't have a clear-cut answer. It will depend on the capabilities of the hardware platform and the details of the library implementation. While conceptually, ldexpf() looks like simple manipulation of the exponent part of a floating-point operand, it is actually a bit more complicated than that, once one considers overflow and gradual underflow via denormals. The latter case involves the rounding of the significand (mantissa) part of the floating-point number.

As ldexp() is generally an infrequently used function, it is in my experience fairly common that less of an optimization effort is applied to it by math library writers than to other math functions.

On some platforms, ldexp(), or a faster (custom) version of it, will be used as a building block in the software implementation of exp2(). The following code provides an exemplary implementation of this approach for float arguments:

#include <cmath>

/* Compute exponential base 2. Maximum ulp error = 0.86770 */
float my_exp2f (float a)
{
    const float cvt = 12582912.0f; // 0x1.8p23
    const float large = 1.70141184e38f; // 0x1.0p127
    float f, r;
    int i;

    // exp2(a) = exp2(i + f); i = rint (a)
    r = (a + cvt) - cvt;
    f = a - r;
    i = (int)r;
    // approximate exp2(f) on interval [-0.5,+0.5]
    r =             1.53720379e-4f;  // 0x1.426000p-13f
    r = fmaf (r, f, 1.33903872e-3f); // 0x1.5f055ep-10f
    r = fmaf (r, f, 9.61817801e-3f); // 0x1.3b2b20p-07f
    r = fmaf (r, f, 5.55036031e-2f); // 0x1.c6af7ep-05f
    r = fmaf (r, f, 2.40226522e-1f); // 0x1.ebfbe2p-03f
    r = fmaf (r, f, 6.93147182e-1f); // 0x1.62e430p-01f
    r = fmaf (r, f, 1.00000000e+0f); // 0x1.000000p+00f
    // exp2(a) = 2**i * exp2(f);
    r = ldexpf (r, i);
    if (!(fabsf (a) < 150.0f)) {
        r = a + a; // handle NaNs
        if (a < 0.0f) r = 0.0f;
        if (a > 0.0f) r = large * large; // + INF
    }
    return r;
}

Most real-life implementations of exp2() do not invoke ldexp(), but a custom version, for example when fast bit-wise transfer between integer and floating-point data is supported, here represented by internal functions __float_as_int() and __int_as_float() that re-interpret an IEEE-754 binary32 as an int32 and vice versa:

/* For a in [0.5, 4), compute a * 2**i, -250 < i < 250 */
float fast_ldexpf (float a, int i)
{
    int ia = (i << 23) + __float_as_int (a); // scale by 2**i
    a = __int_as_float (ia);
    if ((unsigned int)(i + 125) > 250) { // |i| > 125
        i = (i ^ (125 << 23)) - i; // ((i < 0) ? -125 : 125) << 23
        a = __int_as_float (ia - i); // scale by 2**(+/-125)
        a = a * __int_as_float ((127 << 23) + i); // scale by 2**(+/-(i%125))
    }
    return a;
}

On other platforms, the hardware provides a single-precision version of exp2() as a fast hardware instruction. Internal to the processor these are typically implemented by a table lookup with linear or quadratic interpolation. On such hardware platforms, ldexp(float) may be implemented in terms of exp2(float), for example:

float my_ldexpf (float x, int i)
{
    float r, fi, fh, fq, t;

    fi = (float)i;
    /* NaN, Inf, zero require argument pass-through per ISO standard */
    if (!(fabsf (x) <= 3.40282347e+38f) || (x == 0.0f) || (i == 0)) {
        r = x;
    } else if (abs (i) <= 126) {
        r = x * exp2f (fi);
    } else if (abs (i) <= 252) {
        fh = (float)(i / 2);
        r = x * exp2f (fh) * exp2f (fi - fh);
    } else {
        fq = (float)(i / 4);
        t = exp2f (fq);
        r = x * t * t * t * exp2f (fi - 3.0f * fq);
    }
    return r;

}

Lastly, there are platforms that basically provide both exp2() and ldexp() functionality in hardware, such as the x87 instructions F2XM1 and FSCALE on x86 processors.