kupl_egroup_borrow

接口定义

int kupl_egroup_borrow(kupl_egroup_h dest, kupl_egroup_h src);

参数

返回值

• 成功：返回取用executor后的dest egroup大小
• 失败：返回KUPL_ERROR

示例

#include <stdio.h> 
#include "kupl.h" 
   
 int main() 
 { 
         int executor_num = kupl_get_num_executors(); 
         int n1 = executor_num/2; 
         int n2 = executor_num - executor_num/2;
         int executors1[n1], executors2[n2]; 
         for (int i =0; i < n1; i++) {
             executors1[i] = i;
         } 
         for (int i =0; i < n2; i++) {
             executors2[i] = i + n1;
         } 
         kupl_egroup_h egroup1 = kupl_egroup_create(executors1, n1); 
         kupl_egroup_h egroup2 = kupl_egroup_create(executors2, n2); 
         printf("egroup1 : %d executors\n", n1); 
         n1 = kupl_egroup_borrow(egroup1, egroup2); 
         printf("egroup1 : %d executors\n", n1); 
         kupl_egroup_destroy(egroup1); 
         kupl_egroup_destroy(egroup2); 
         return 0; 
 }

运行结果如下。

egroup1 : 2 executors
egroup1 : 4 executors

上述示例演示了创建的egroup1取用egroup2中所有的executor的流程，运行结果打印了取用前与取用后egroup1中的executor数量，由此可见kupl_egroup_borrow函数使egroup1取用了egroup2的所有executors。