《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;
}
}