《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 
   
    #include 
    
     #include 
     
      #include 
      
       #include 
       
        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 
   
    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 
   
    #include 
    
     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;  }}