问题描述
我必须解决有关OpenMP的练习；在这里:
I have to solve an exercise about OpenMP; here it is:
对于最终的计算，我将使用
For the final calculation I would use
#pragma omp parallel for shared(A,B)
for(int i=0; i<NUM; i++) {
B[i] = A[i]/MaxA;
}
我的疑问在于如何利用OpenMP来计算A的最大值.
My doubt is on how to take advantage of OpenMP to compute the maximum value of A.
我想到的唯一想法是使用并行部分并行化递归最大计算:
The only idea I came up with is using parallel sections to parallelize a recursive maximum computation:
第一个呼叫是i = 0；j = sizeOfA-1
The first call is made with i=0; j= sizeOfA - 1
int max(int A[], int i, int j) {
// Leaves conditions
switch (j-i) {
case 1:
{ if( A[i]>A[j] )
return A[i];
else
return A[j];
}
break;
case 0:
return A[i];
break;
}
int left, right;
#pragma omp parallel sections
{
#pragma omp section
{
left = max( A, i, i+(j-i)*0.5);
}
#pragma omp section
{
right = max( A, i+(j-i)*0.5+1, j);
}
}
// Nodes conditions
if( right > left )
return right;
else
return left;
}
您认为这是一个很好的解决方案吗?你能告诉我是否有更好的解决方案/替代方法?
Do you think it is a good solution? Can you tell me if there is any better solution / alternative ?
推荐答案
如何使用归约子句来计算[i，j]之间范围内A内的最大值，而不是使用递归计算?
What about using a reduction clause to calculate the maximum value within A in the range between [i,j] instead of using recursive calculation?
类似
int max(int A[], int i, int j)
{
int idx;
int max_val; /* = 0 not needed according to Jim Cownie comment */
#pragma omp parallel for reduction(max:max_val)
for (idx = i; idx < j; idx++)
max_val = max_val > A[idx] ? max_val : A[idx];
return max_val;
}
这篇关于使用OpenMP的数组中的最大值的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！