我正在尝试制作一个使用效率非常低的算法的程序,该算法使用 POSIX 线程计算范围内的完美数。我似乎无法很好地掌握锁定的概念以使我的算法正常工作。我想返回一个完美数字列表。任何人都可以就如何更好地实施这一点提供一些建议吗?
具体问题: - 我如何让它只打印出每个完美数字的 1 个瞬间?- 如何让它返回值而不是只打印值?
资源:
static pthread_mutex_t mtx = PTHREAD_MUTEX_INITIALIZER;
static void * threadFunc(void *arg) {
int range, start, end;
int i, s, number, mod, divisor, temp, sum;
s = pthread_mutex_lock(&mtx);
/* Takes in a string and pulls out the two integers */
sscanf(arg,"%i-%i", &start, &end);
printf("\nStart: %i\nEnd: %i\n", start, end);
printf("\n");
s = pthread_mutex_unlock(&mtx);
for (number=start; number<=end; number++) { // loop through range of numbers
temp=0,sum=0;
// loops through divisors of a number and sums up whole divisors
for (i=1; i<number; i++) {
//s = pthread_mutex_lock(&mtx);
mod = number % i;
//s = pthread_mutex_unlock(&mtx);
if (mod == 0){
s = pthread_mutex_lock(&mtx);
divisor = i;
sum = divisor + temp;
temp = sum;
s = pthread_mutex_unlock(&mtx);
}
}
//if the sum of whole divisors is equal to the number, its perfect
if (sum == number) {
s = pthread_mutex_lock(&mtx);
printf("%i is a Perfect Number \n", sum);
//return sum somehow;
s = pthread_mutex_unlock(&mtx);
}
}
return NULL;
}
int main(int argc, char *argv[]) {
pthread_t tid[5];
int prefect_number, i, s;
char input[]="1-9999";
for(i=0; i < 5; ++i) {
pthread_create(&tid[i], NULL, &threadFunc, input);
print_thread_info();
}
/* Wait for the perfect number thread to complete, then get result. */
for(i=0; i < 5; ++i)
pthread_join(tid[i],NULL);
return 0;
}