[第五章] 深入理解计算机系统第三版 家庭作业参考答案

5.13

A.
画图：

关键路径为第三幅图加粗部分

B.
下界为浮点加法的延迟界限，CPE 为 3.00

C.
整数加法的延迟界限，CPE 为 1.00

D.
关键路径上只有浮点加法

5.14

void inner5(vec_ptr u, vec_ptr v, data_t *dest) {
	long i;
	long length = vec_length(u);
	long limit = length - 5;
	data_t *udata = get_ver_start(u);
	data_t *vdata = get_vec_start(v);
	data_t sum = (data_t)0;

	for (i = 0; i < limit; i += 6) {
		sum = sum + udata[i] * vdata[i] + 
			udata[i + 1] * vdata[i + 1] + 
			udata[i + 2] * vdata[i + 2] + 
			udata[i + 3] * vdata[i + 3] + 
			udata[i + 4] * vdata[i + 4] + 
			udata[i + 5] * vdata[i + 5];
	}

	for (; i < length; i++) {
		sum = sum + udata[i] * vdata[i];
	}

	*dest = sum;
}

A.
之所以任何标量版本都无法达到比 1.00 小的 CPE，是因为每个时钟周期只能加载两个值；即使流水线是满的，CPE 也只能为 1.00
B.
无论循环展开多少次，关键路径上还是有 length 个浮点加法

5.15

void inner6(vec_ptr u, vec_ptr v, data_t *dest) {
	long i;
	long length = vec_length(u);
	long limit = length - 5;
	data_t *udata = get_ver_start(u);
	data_t *vdata = get_vec_start(v);
	data_t sum0 = (data_t)0;
	data_t sum1 = (data_t)0;
	data_t sum2 = (data_t)0;
	data_t sum3 = (data_t)0;
	data_t sum4 = (data_t)0;
	data_t sum5 = (data_t)0;

	for (i = 0; i < limit; i += 6) {
		sum0 = sum0 + udata[i] * vdata[i];
		sum1 = sum1 + udata[i + 1] * vdata[i + 1];
		sum2 = sum2 + udata[i + 2] * vdata[i + 2];
		sum3 = sum3 + udata[i + 3] * vdata[i + 3];
		sum4 = sum4 + udata[i + 4] * vdata[i + 4];
		sum5 = sum5 + udata[i + 5] * vdata[i + 5];
	}

	for (; i < length; i++) {
		sum0 = sum0 + udata[i] * vdata[i];
	}

	*dest = sum0 + sum1 + sum2 + sum3 + sum4 + sum5;
}

只有两个加载单元，一个时钟周期只能加载两个值，CPE 最低只能到 1.00

5.16

void inner7(vec_ptr u, vec_ptr v, data_t *dest) {
	long i;
	long length = vec_length(u);
	long limit = length - 5;
	data_t *udata = get_ver_start(u);
	data_t *vdata = get_vec_start(v);
	data_t sum = (data_t)0;

	for (i = 0; i < limit; i += 6) {
		sum = sum + (udata[i] * vdata[i] +
			(udata[i + 1] * vdata[i + 1] +
			(udata[i + 2] * vdata[i + 2] +
			(udata[i + 3] * vdata[i + 3] +
			(udata[i + 4] * vdata[i + 4] +
			udata[i + 5] * vdata[i + 5])))));
	}

	for (; i < length; i++) {
		sum = sum + udata[i] * vdata[i];
	}

	*dest = sum;
}

5.17

void *new_memset(void *s, int c, size_t n) {
	unsigned long w;
	unsigned char *pw = (unsigned char *)&w;
	size_t cnt = 0;
	while (cnt < K) {
		*pw++ = (unsigned char)c;
		cnt++;
	}
	
	size_t i;
	unsigned char *schar = s;
	for (i = 0; (size_t)schar % K != 0 || i == n; i++) {
		*schar++ = (unsigned char)c;
	}

	size_t limit = n - K + 1;
	for (; i < limit && (int)limit > 0; i += K) {
		*(unsigned long *)schar = w;
		schar += K;
	}

	for (; i < n; i++) {
		*schar++ = (unsigned char)c;
	}

	return s;
}

5.18

6*3a

double poly_6_3a(double a[], double x, long degree) {
  long i = 1;
  double result = a[0];
  double result1 = 0;
  double result2 = 0;

  double xpwr = x;
  double xpwr1 = x * x * x;
  double xpwr2 = x * x * x * x * x;

  double xpwr_step = x * x * x * x * x * x;
  for (; i < degree - 5; i+=6) {
    result = result + (a[i]*xpwr + a[i+1]*xpwr*x);
    result1 = result1 + (a[i+2]*xpwr1 + a[i+3]*xpwr1*x);
    result2 = result2 + (a[i+4]*xpwr2 + a[i+5]*xpwr2*x);

    xpwr *= xpwr_step;
    xpwr1 *= xpwr_step;
    xpwr2 *= xpwr_step;
  }

  for (; i <= degree; i++) {
    result = result + a[i]*xpwr;
    xpwr *= x;
  }

  return result + result1 + result2;
}

5.19

void psum_4_1a(float a[], float p[], long n) {
	long i;
	float tmp, tmp1, tmp2, tmp3 = 0;

	for (i = 0; i < n - 3; i += 4) {
		tmp = tmp3 + a[i];
		tmp1 = tmp + a[i + 1];
		tmp2 = tmp1 + a[i + 2];
		tmp3 = tmp2 + a[i + 3];

		p[i] = tmp;
		p[i + 1] = tmp1;
		p[i + 2] = tmp2;
		p[i + 3] = tmp3;
	}

	for (; i < n; i++) {
		tmp3 += a[i];
		p[i] = tmp3;
	}
}