pthread 函数的返回值与函数内的计算值不同

Question

现在我正在尝试使用 pthread 函数 return 值。在下面的代码中，我想得到的输出是“38 99”或“99 38”，但程序输出“99 99”或“38 38”。当我把 "cout" 放在 pthread 函数中时，输出是正确的，但是当我把 "cout" 放在 main 函数中时，输出是错误的。谁能帮我看看哪里做错了？

using namespace std;

#define NTHREADS 2
pthread_mutex_t mutex1 = PTHREAD_MUTEX_INITIALIZER;

struct args{
  int * numbers;
  float result;
};

void *add(void* a){
   struct args *number = (struct args *)a;
   pthread_mutex_lock( &mutex1 );
   int *n = (int*) number->numbers;
   float sum =0;
   for(int i = 0; i < 5; i++){
     sum = sum + n[i] +5;
   }
   float av = sum/5;
   number-> result  = av;
   //cout << number->result <<endl;
    pthread_mutex_unlock( &mutex1 );
   return number;  
   }

void *substract(void* a){
 struct args *number = (struct args *)a;
 pthread_mutex_lock( &mutex1 );
 int *n = (int*) number->numbers;
 float sum =0;
 for(int i = 0; i < 5; i++){
    sum = sum + n[i] *3;
 }
 float av = sum/5;
 number-> result  = av;
 //  cout << number->result <<endl;
 pthread_mutex_unlock( &mutex1 );

 return number;
}

main(){
pthread_t thread_id[NTHREADS];
int i, j;

int *numbers = new int[5];
numbers[0] = 34; numbers[1] = 2; numbers[2]= 77; numbers[3] = 40; numbers[4] = 12;

struct args a;
a.numbers = numbers;
pthread_create( &thread_id[0], NULL, add, (void*) &a);
pthread_create( &thread_id[1], NULL, substract, (void*) &a );
void *status1;
void * status2;

pthread_join( thread_id[0], &status1);
pthread_join( thread_id[1], &status2);
args status = *(args*) status1;
float ra = status.result;
cout << ra << endl;

args statuss = *(args*) status2;
float rb = statuss.result;
cout << rb << endl;

 exit(EXIT_SUCCESS);
}

Answer 1

你得到这个输出是因为两个线程都将它们的结果写入同一个地方 - struct args a.

需要在main函数中分配两个 struct args，将一个的地址传给第一个线程，另一个传给第二个线程。

pthread 函数的返回值与函数内的计算值不同

Returned value from pthread function is different from the calculated value within function

c++

pthreads