我对这段代码有一个非常奇怪的问题，抱歉它很乱。基本上它是一个pagerank算法。每个结构网页都包含在动态数组“pages”中。pages向量通过算法直到它的绝对值（| P|) 小于 'epsilon'。现在问题出在第 195-201 行。如果我在这些行中删除对数组的迭代（即一个空的while循环），它适用于只需要一次迭代的情况。但是，当我确实有 for 循环（即使是一次迭代情况）时，它会抛出 error6（第 179 行，调试显示 e == NULL），甚至没有运行插入的循环。我已经设置了断点等，甚至在没有阅读额外代码的情况下仍然给出错误6。这里发生了什么？我对 C 和并行编程很陌生，所以它可能是基本的东西。将不胜感激任何帮助！
输入格式：
number_of_cores
number_of_pages
...
page_names
...
page_links
输出格式：
...
page_rank
...
代码
#include <assert.h>
#include <math.h>
#include <pthread.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
static const double D = 0.85;
static const double EPSILON = 0.005;
int ncores;
int npages;
struct webpage** pages;
int maxdepth;
struct webpage* has(char s[20], int e);
void* threadf(void* ptr);
int quit(void);
double rec(int s, int f, int depth);
struct webpage {
char name[20];
double oldrank;
double rank;
struct node* in;
int incount;
int outcount;
};
struct node {
struct webpage* data;
struct node* next;
};
struct arg {
int s;
int f;
int depth;
double ret;
};
struct webpage*
has(char s[20], int e) {
int p;
for (p=0; p<e; ++p) {
if (strcmp(s, pages[p]->name) == 0) {
return pages[p];
}
}
return NULL;
}
void *
threadf(void* ptr) {
struct arg* curr = (struct arg*)ptr;
curr->ret = rec(curr->s, curr->f, curr->depth);
}
int
quit(void) {
int i;
for(i=0; i<npages; ++i) {
struct node* curr = pages[i]->in;
struct node* next;
while(curr != NULL) {
next = curr->next;
free(curr);
curr = next;
}
free(pages[i]);
}
free(pages);
return 0;
}
double
seq(int s, int f) {
double sum;
sum = 0;
int w;
for (w=s; w<=f; w++) {
struct webpage* curr = pages[w];
double ser;
ser = 0;
struct node* currn = curr->in;
while (currn != NULL) {
struct webpage* n = currn->data;
ser = ser + ((n->oldrank)/(n->outcount));
currn = currn->next;
}
double temp = (((1-D)/npages) + (D*ser));
sum = sum + pow((temp - curr->oldrank), 2);
curr->oldrank = curr->rank;
curr->rank = temp;
}
return sum;
}
double
rec(int s, int f, int depth) {
if (depth == maxdepth ) {
return seq(s, f);
} else {
if (s < f){
int m;
m = (s+f)/2;
struct arg l;
struct arg r;
l.s = s;
l.f = m;
l.depth = depth+1;
r.s = m+1;
r.f = f;
r.depth = depth+1;
pthread_t left, right;
pthread_create(&left, NULL, threadf, (void*) &l);
pthread_create(&right, NULL, threadf, (void*) &r);
pthread_join(left, NULL);
pthread_join(right, NULL);
double res;
res = l.ret + r.ret;
return res;
}
return seq(s, f);
}
}
int
main(void) {
if (scanf("%d", &ncores) != 1) {
printf("error1\n");
return quit();
}
if (scanf(" %d", &npages) != 1) {
printf("error2\n");
return quit();
}
int i;
char n[20];
pages = (struct webpage**)malloc(npages*sizeof(struct webpage*));
for (i=0; i<npages; ++i) {
if (scanf(" %c", n) != 1 || has(n, i) != NULL) {
printf("error3\n");
return quit();
}
pages[i] = (struct webpage*)malloc(sizeof(struct webpage));
struct webpage* curr = pages[i];
strcpy(curr->name, n);
curr->oldrank = 1/npages;
curr->in = NULL;
curr->incount = 0;
curr->outcount = 0;
}
int nedges;
if (scanf(" %d", &nedges) != 1) {
printf("error4\n");
return quit();
}
for (i=0; i<nedges; ++i) {
char f[20], t[20];
if (scanf(" %s %s", f, t) != 2) {
printf("error5\n");
return quit();
}
char from[20], to[20];
strcpy(from, f);
strcpy(to, t);
struct webpage* s = has(from, npages);
struct webpage* e = has(to, npages);
if (s == NULL || e == NULL) {
printf("error6\n");
return quit();
}
s->outcount++;
e->incount++;
struct node* new;
new = (struct node*)malloc(sizeof(struct node));
new->data = s;
if (e->in == NULL) {
e->in = new;
} else {
new->next = e->in;
e->in = new;
}
}
maxdepth = (log(ncores))/(log(2)) + 0.5;
while (sqrt(rec(0, npages-1, 0)) > EPSILON){
int c;
for (c=0; c<npages; ++c) {
struct webpage* curr = pages[c];
curr->oldrank = curr->rank;
}
}
int z;
for (z=0; z<npages; ++z) {
struct webpage* curr = pages[z];
printf("%s %.4lf\n", curr->name, curr->rank);
}
return quit();
}
样本输入：
8
4
a
b
c
d
4
a a
输出：
error6