我有一个函数，它计算一个较大字符串的数字子字符串（使用一个较小的字符串进行匹配），我试图用pthreads替换这个序列，从而在没有循环的情况下同时处理整个事情。
我在寻找我能做到这一点的原则。
到目前为止，我所做的是创建一个pthread_tdynamic数组，这个数组的数量与字符的数量相同。在更大的字符串中，我将计算子字符串的函数分配给线程，我想我几乎消化了它，我只需要向前推一推就可以得到正确的结果。
pthreads.c文件
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <pthread.h>
#define MAX 1024
int total = 0;
int n1,n2;
int num_thr;
int *num_threads;
char *s1,*s2;
FILE *fp;
int readf(FILE *fp)
{
if((fp=fopen("strings.txt", "r"))==NULL){
printf("ERROR: can't open string.txt!\n");
return 0;
}
s1=(char *)malloc(sizeof(char)*MAX);
if(s1==NULL){
printf("ERROR: Out of memory!\n");
return -1;
}
s2=(char *)malloc(sizeof(char)*MAX);
if(s1==NULL){
printf("ERROR: Out of memory\n");
return -1;
}
/*read s1 s2 from the file*/
s1=fgets(s1, MAX, fp);
s2=fgets(s2, MAX, fp);
n1=strlen(s1)-1; /*length of s1*/
n2=strlen(s2)-1; /*length of s2*/
num_thr=n1;
printf("String 1 len = %d\n",n1);
if(s1==NULL || s2==NULL || n1<n2) /*when error exit*/
return -1;
}
void *num_substring(void *vari);
void *num_substring(void *vari)
{
int i,j,k;
int count;
for (i = 0; i <= (n1-n2); i++){
count=0;
for(j = i,k = 0; k < n2; j++,k++){ /*search for the next string of size of n2*/
if (*(s1+j)!=*(s2+k)){
break;
}
else
count++;
if(count==n2)
total++; /*find a substring in this step*/
}
}
printf("total= %d\n",total);
//return total;
}
int main (int argc, char *argv[])
{
int count;
readf(fp);
printf("The number of substrings is: %d\n", count);
int i, ret=-1;
char *msg1= "a thread";
pthread_t * thread_arr = malloc(sizeof(pthread_t)*num_thr);
printf("num threads inside main = %d\n",num_thr);
for (i = 0; i < num_thr; i++) {
ret = pthread_create(&thread_arr[i], NULL, num_substring, (void *) msg1);
if(ret != 0) {
printf ("Create pthread %d error!\n",i);
exit (1);
}
printf("Main function thread %d created\n",i);
}
for (i=0;i < num_thr; i++){
pthread_join(thread_arr[i], NULL);
}
return 0;
}
此程序的正确输出必须为4，因为
ab在另一个字符串中出现4次，因此子字符串的数目必须等于4：string.txt内容：
阿不都木
ab公司
我的输出不是完全错误的，因为它最初显示4，然后它开始递增（这不是我想要的全部是4），但是，它仍然没有完成，因为我没有用pthread s array（pthread_arr）替换循环，这是我猜测需要的最后一步，这里是我的输出：
String 1 len = 22
The number of substrings is: 10219508
num threads inside main = 22
Main function thread 0 created
Main function thread 1 created
Main function thread 2 created
Main function thread 3 created
total= 4 // this is the correct number of substrings
total= 8
total= 12
Main function thread 4 created
total= 20
total= 16
Main function thread 5 created
Main function thread 6 created
total= 24
Main function thread 7 created
total= 28
Main function thread 8 created
total= 36
total= 32
Main function thread 9 created
total= 40
Main function thread 10 created
total= 44
Main function thread 11 created
Main function thread 12 created
total= 48
total= 52
Main function thread 13 created
total= 56
Main function thread 14 created
total= 60
Main function thread 15 created
Main function thread 16 created
Main function thread 17 created
Main function thread 18 created
Main function thread 19 created
Main function thread 20 created
Main function thread 21 created
total= 64
total= 68
total= 84
total= 76
total= 72
total= 80
total= 88
最佳答案
你离得很近，但有一些虫子。
每个线程都在扫描整个s1而不仅仅是一个子字符串。main应传递一个参数，该参数指示s1中子字符串的起始偏移量
线程函数中只需要一个循环。
每个线程递增一个全局total[没有线程锁定]。最好将值作为线程的返回值传回
因为s1中每个字符都有一个线程，所以线程将0或1视为计数[仅限]。
我已经修正了你的代码[请原谅无端的样式清理]：
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <pthread.h>
#define MAX 1024
int n1;
int n2;
int num_thr;
int *num_threads;
char *s1;
char *s2;
FILE *fp;
int
readf(FILE * fp)
{
char *cp;
if ((fp = fopen("strings.txt", "r")) == NULL) {
printf("ERROR: can't open string.txt!\n");
return 0;
}
if (s1 == NULL)
s1 = (char *) malloc(sizeof(char) * MAX);
if (s1 == NULL) {
printf("ERROR: Out of memory!\n");
return -1;
}
if (s2 == NULL)
s2 = (char *) malloc(sizeof(char) * MAX);
if (s1 == NULL) {
printf("ERROR: Out of memory\n");
return -1;
}
/* read s1 s2 from the file */
cp = fgets(s1, MAX, fp);
if (cp == NULL)
return -1;
cp = fgets(s2, MAX, fp);
if (cp == NULL)
return -1;
n1 = strlen(s1) - 1; /* length of s1 */
n2 = strlen(s2) - 1; /* length of s2 */
num_thr = n1;
printf("String 1 len = %d\n", n1);
if (n1 < n2)
return -1;
return 0;
}
void *
num_substring(void *vari)
{
int i;
long subtotal;
subtotal = 0;
#if 0
int count;
int j;
int k;
for (int i = 0; i <= (n1 - n2); i++) {
count = 0;
/* search for the next string of size of n2 */
for (int j = i, k = 0; k < n2; j++, k++) {
if (*(s1 + j) != *(s2 + k)) {
break;
}
else
count++;
if (count == n2)
subtotal++; /* find a substring in this step */
}
}
#else
int t = (long) vari;
char *cp = &s1[t];
subtotal = 1;
for (i = 0; i < n2; i++) {
//printf("t%d: TRY: %c %c\n",t,cp[i],s2[i]);
if (cp[i] != s2[i]) {
subtotal = 0;
break;
}
}
#endif
//printf("t%d: subtotal= %ld\n", subtotal);
return (void *) subtotal;
}
int
main(int argc, char *argv[])
{
int total = 0;
void *ptr;
//int count;
readf(fp);
//printf("The number of substrings is: %d\n", count);
long i;
int ret = -1;
//char *msg1 = "a thread";
pthread_t *thread_arr = malloc(sizeof(pthread_t) * num_thr);
printf("num threads inside main = %d\n", num_thr);
for (i = 0; i < num_thr; i++) {
#if 0
ret = pthread_create(&thread_arr[i], NULL, num_substring,(void *) msg1);
#else
ret = pthread_create(&thread_arr[i], NULL, num_substring,(void *) i);
#endif
if (ret != 0) {
printf("Create pthread %ld error!\n", i);
exit(1);
}
printf("Main function thread %ld created\n", i);
}
total = 0;
for (i = 0; i < num_thr; i++) {
pthread_join(thread_arr[i], &ptr);
total += (long) ptr;
}
printf("TOTAL: %d\n",total);
return 0;
}
关于c - 如何在C中用pthread替换顺序操作？，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/42537855/