Subtracting an integer comparison result with SSE

Question

I'm learning SSE coding and now stuck on a couple lines of code while porting / vectorizing an exp(double) function.

I'm kind of beginner with C/C++ coding as well so these two lines confuses and I don't know what's happening there:

fast_exp(double x) {
    double  a = c1 * x + 0.5;
    int     n = (int)a;  // truncate toward zero
    ieee754 u;  // union of double and unsigned short[4]

    n -= (a < 0); //(1)
     ...

    u.s[3] = (unsigned short)((n << 4) & 0x7FF0); //(2)
    return  u.d;
}

What is done there?

(1) Is the a subtracted from n if a is holding a negative value ?

(2) ? bitwise left shift but, what more happens there?

And, how those two code lines are written using SSE intrinsics?

---

Here's what I have so far, partially ported to SSE intrinsics. The lines I'm stuck on are marked with ?? in code below. It doesn't compile (yet).

typedef union {
  double d;
  unsigned short s[4];
} ieee754;

double exp_sse (double value){

    double px; //, a;
    ieee754 u;
    __m128i n;
    __m128d a;
    __m128d  x = _mm_set1_pd (x);
    __m128d c1 = _mm_set1_pd (1.4426950408889634073599);
    __m128d c2 = _mm_set1_pd (6.93145751953125E-1);
    __m128d c3 = _mm_set1_pd (1.42860682030941723212E-6);
    __m128i c1023 = _mm_set1_epi32(1023);
    __m128d c4 = _mm_set1_pd (0.5);

    /* n = round(x / log 2) --------------------- */
    // a = c1 * x + 0.5; 
    a = _mm_mul_pd (c1, x);
    a = _mm_add_pd (a, c4);

    // n = (int)a; 
    n = _mm_cvtpd_epi32(a);

    n -= (a < 0); ?? 

    /* x -= n * log2 ---------------------- */
    //px = (double)n;
    px = _mm_cvtepi32_pd(n);
    //x -= px * c2;
    x = _mm_sub_pd(x, _mm_mul_pd(px, c2));
    //x -= px * c3;
    x = _mm_sub_pd(x, _mm_mul_pd(px, c3));

    // calc e^x -------------------
       ...
    // ----------------------------

    /* 2^n in double. */
    n = _mm_add_(n, c1023);

    u.s[3] = (unsigned short)((n << 4) & 0x7FF0); ??

    return d[0] * u.d;
}