《CSAPP》（第3版）答案（第五章）

《CSAPP》（第3版）答案（第五章）

P13

• A
  图片来源：
  https://github.com/DreamAndDead/CSAPP-3e-Solutions/blob/master/chapter5/5.13.md
  
  
• B
  3.0
• C
  1.0
• D
  只在关键路径上进行浮点数加法

P14

• A
  每个元素需要六次长整型或浮点数加法，一共n/6个元素，所以是6*n/6=n次长整型或浮点数加法。CPE下限1.0
• B
  与A同理

P15

void inner4(vec_ptr u, vec_ptr v, data_t *dest) {
  long i;
  long length = vec_length(u);
  data_t *udata = get_vec_start(u);
  data_t *vdata = get_vec_start(v);
  data_t sum = (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 < length-6; i+=6) {
    sum = sum + 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++) {
    sum = sum + udata[i] * vdata[i];
  }
  *dest = sum + sum1 + sum2 + sum3 + sum4 + sum5;
}

限制性能的因素：我也不知道

P16

void inner4(vec_ptr u, vec_ptr v, data_t *dest) {
  long i;
  long length = vec_length(u);
  data_t *udata = get_vec_start(u);
  data_t *vdata = get_vec_start(v);
  data_t sum = (data_t) 0;
  
  for (i = 0; i < length-6; 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;
}

P17

#include <stdio.h>
#include <stdint.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>

void* basic_memset(void *s, int c, size_t n) {
  size_t cnt = 0;
  unsigned char *schar = s;
  while (cnt < n) {
    *schar++ = (unsigned char) c;
    cnt++;
  }
  return s;
}

void* effective_memset(void *s, unsigned long cs, size_t n) {
  size_t K = sizeof(unsigned long);
  size_t cnt = 0;
  unsigned char *schar = s;
  while (cnt < n) {
    if ((size_t)schar % K == 0) {
      break;
    }
    *schar++ = (unsigned char)cs;
    cnt++;
  }
  
  unsigned long *slong = (unsigned long *)schar;
  size_t rest = n - cnt;
  size_t loop = rest / K;
  size_t tail = rest % K;
  for (size_t i = 0; i < loop; i++) {
    *slong++ = cs;
  }
  
  schar = (unsigned char *)slong;
  for (size_t i = 0; i < tail; i++) {
    *schar++ = (unsigned char)cs;
  }
  return s;
}

P18

#include <stdio.h>
double poly(double a[], double x, long degree) {
  long i;
  double result = a[0];
  double xpwr = x;
  for (i = 1; i <= degree; i++) {
    result += a[i] * xpwr;
    xpwr = x * xpwr;
  }
  return result;
}

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 - 6; 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;
}

double polyh(double a[], double x, long degree) {
  long i;
  double result = a[degree];
  for (i = degree-1; i >= 0; i--) {
    result = a[i] + x*result;
  }
  return result;
}

P19

#include <stdio.h>
#include <assert.h>

void psum1a(float a[], float p[], long n) {
  long i;
  float last_val, val;
  last_val = p[0] = a[0];
  for (i = 1; i < n; i++) {
    val = last_val + a[i];
    p[i] = val;
    last_val = val;
  }
}

void psum_4_1a(float a[], float p[], long n) {
  long i;
  float val, last_val;
  float tmp, tmp1, tmp2, tmp3;
  last_val = p[0] = a[0];
  for (i = 1; i < n - 4; i++) {
    tmp = last_val + 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;
    last_val = last_val + (a[i] + a[i+1] + a[i+2] + a[i+3]);
  }
  for (; i < n; i++) {
    last_val += a[i];
    p[i] = last_val;
  }
}

第五章 完