为什么！=运算符不使用OpenMP允许？ [英] Why is the != operator not allowed with OpenMP?

问题描述

我试图编译以下code：

I was trying to compiled the following code:

#pragma omp parallel shared (j)
{
   #pragma omp for schedule(dynamic)
   for(i = 0; i != j; i++)
   {
    // do something
   }
}

我得到这个错误：错误：无效的控股predicate

我检查的OpenMP 参考指南和它说，对于平行它仅仅允许以下运营商之一：其中， ＆LT; =>> =。

I check the openMP reference guide and it says that for the parallel for it "only" allows one of the following operators: < <= > >=.

我不明白为什么不能让我！= j的。我还可以理解，如果它是静态调度中，由于OPENMP需要$ P $对计算分配给每个线程的迭代次数。但我不明白为什么这种限制在这样的情况下，例如。任何线索？

I don't understand why not allow i != j. I could understand if it was the static schedule, since openMP need to pre-compute the number of iterations assigned to each thread. But i can't understand why this limitation in such case for example. Any clues ?

编辑：就算我做的（！I = 0; i = 100;我++），虽然我可能只是把＆LT;或＆LT; =

even if I make for(i = 0; i != 100; i++), although I could just put "<" or "<=" .

推荐答案

有关符号整数，围绕行为包是不确定的。如果我们允许！=，程序员可能会得到意外tripcount。问题是编译器是否能产生code来计算循环之旅计数。

For signed int, the wrap around behavior is undefined. If we allow !=, programmers may get unexpected tripcount. The problem is whether the compiler can generate code to compute a trip count for the loop.

对于一个简单的循环，如：

For a simple loop, like:

for( i = 0; i < n; ++i )

编译器能够确定有N次迭代，如果n> = 0 ，然后反复零的如果n＆LT; 0

对于这样一个循环：

for( i = 0; i != n; ++i ) 

再次，编译器应该能够确定有N次迭代，如果n> = 0 ; 如果n＆LT; 0 ，我们不知道有多少次迭代了。

again, a compiler should be able to determine that there are 'n' iterations, if n >= 0; if n < 0, we don't know how many iterations it has.

对于这样一个循环：

for( i = 0; i < n; i += 2 )

编译器可以生成code来计算行程计数（循环迭代计数）为楼（（N + 1）/ 2）如果n> = 0 和0 如果n＆LT; 0

对于这样一个循环：

for( i = 0; i != n; i += 2 )

编译器不能确定是否'我'都不会打'N'。如果n是什么奇数？

the compiler can't determine whether 'i' will ever hit 'n'. What if 'n' is an odd number?

对于这样一个循环：

for( i = 0; i < n; i += k )

编译器可以生成code来计算行程计数为楼（（N + K-1）/ K）如果n> = 0 0 若n＆LT ; 0 ，因为编译器知道该回路必须计数;在这种情况下，如果的 K＆下; 0 ，它不是一个合法的OpenMP程序。

the compiler can generate code to compute the trip count as floor((n+k-1)/k) if n >= 0, and 0 if n < 0, because the compiler knows that the loop must count up; in this case, if k < 0, it's not a legal OpenMP program.

对于这样一个循环：

for( i = 0; i != n; i += k )

编译器甚至不知道我是否是或计数了。它不知道'我'都不会打'N'。它可能是一个无限循环

the compiler doesn't even know if i is counting up or down. It doesn't know if 'i' will ever hit 'n'. It may be an infinite loop.

Credites ：OpenMP的ARB

Credites: OpenMP ARB

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