我一直在墙上敲了几个小时,现在试图找出导致这个段错误的原因。
我发现在pthread_mutex_lock(lock)
行(38)上一直发生段错误。我已经在锁定周围放置了两个打印语句,但只打印了其中一个,这是我得出结论是在该指令处发生段错误的理由。
我是否正确使用了互斥锁,或者我的数组(buffer[]
和numbermarker[]
是否犯了错误?
#include <stdio.h>
#include <stdlib.h>
#include <pthread.h>
int* numbermarker = NULL;
int* buffer = NULL;
int pullposition = 0;
int placeposition = 1;
pthread_mutex_t *lock;
int ceiling;
/*This method places one of the primes in the buffer. It
offers a safe way to manage where the next value will be placed*/
void placevalue(int value)
{
buffer[placeposition] = value;
placeposition++;
}
/*This method pulls the next prime and increments to the next prime in the list*/
int takevalue()
{
pullposition++;
return buffer[pullposition-1];
}
void* threadmethod()
{
int k;
int l;
int firstval;
while(1)
{
while(numbermarker[buffer[pullposition]-1]==0)
{
printf("flag1 \n");
pthread_mutex_lock(lock);
printf("flag2 \n");
numbermarker[buffer[pullposition]-1] = 1;
l = takevalue();
pthread_mutex_unlock(lock);
firstval = 1;
for(k=l+1;k<=ceiling;k++)
{
if(k%l != 0)
{
if(firstval)
{
placevalue(k);
firstval = 0;
}
}
else
{
numbermarker[k-1] = 1;
}
}
}
}
}
int main()
{
int numthreads;
int i;
printf("Enter number of threads: \n");
scanf("%d", &numthreads);
printf("Enter the highest value to check \n");
scanf("%d", &ceiling);
/* This will hold 1's and 0's.
1 = number has been checked or is
confirmed not to be a prime
0 = number is a possible prime
The idea behind these values is that the next
prime can always be identified by the 0 with
the lowest index*/
numbermarker = (int*)malloc(sizeof(int)*(ceiling));
/*This will hold the primes as they are found*/
buffer = (int*)malloc(sizeof(int)*(ceiling));
for(i=0; i<ceiling; i++)
{
if(i<1)
{
numbermarker[i] = 1;
}
else
{
numbermarker[i] = 0;
}
buffer[i]=0;
printf("%d \n",numbermarker[i]);
}
placevalue(2);
pthread_t **tid = (pthread_t **) malloc(sizeof(pthread_t *) * numthreads);
for(i=0;i<numthreads;i++)
{
tid[i] = (pthread_t *) malloc(sizeof(pthread_t));
}
for(i=0;i<numthreads;i++)
{
if(pthread_create(tid[i],
NULL,
threadmethod,
NULL))
{
printf("Could not create thread \n");
exit(-1);
}
}
int not_done = 1;
int sum;
while(not_done)
{
sum = 0;
for(i=0; i<ceiling; i++)
{
sum += numbermarker[i];
}
if(sum == ceiling)
not_done = 0;
}
for(i=0;i<numthreads;i++)
{
if(pthread_join(*tid[i], NULL))
{
printf("Error Joining with thread \n");
}
free(tid[i]);
}
free(tid);
for(i=0;i<ceiling;i++)
{
if(buffer[i] != 0);
printf("%d \n", i);
}
free(buffer);
free(numbermarker);
buffer=NULL;
numbermarker=NULL;
return(0);
}
答案 0 :(得分:5)
lock
是一个未初始化的指针。您需要分配内存,然后在锁定之前初始化互斥锁。最简单的解决方法是更改
pthread_mutex_t *lock;
到
pthread_mutex_t lock = PTHREAD_MUTEX_INITIALIZER;
和(因为lock
不再是指针)
pthread_mutex_lock(lock);
....
pthread_mutex_unlock(lock);
到
pthread_mutex_lock(&lock);
....
pthread_mutex_unlock(&lock);