reduce_parallel 是不是线程安全函数?
reduce_parallel is not a thread-safe function?
我想调用parallel_reduce对向量元素求和。但是我发现如果向量元素足够了,结果是不正确的。请帮助我如何使用此功能。
// prepare data
const size_t allNum = 1000000;
std::vector<double> a;
for (int i = 0; i < allNum; ++i)
{
a.push_back(double(i + 1));
}
// λ func
auto f = [&]() -> double {
return tbb::parallel_reduce(tbb::blocked_range<size_t>(0, allNum),
0.0,
[&](const tbb::blocked_range<size_t>& r, double init) -> double {
for (int i = r.begin(); i < r.end(); ++i)
{
init += a[i];
}
return init;
},
[](double f, double s) -> double {
return f + s;
}
/*std::plus<double>()*/);
};
// call λ func, get the result
double correctResult = (1.0 + 1000000.0) * 500000.0;
double sum = f(); // sum != correctResult
// sum is different every loop
我试过运行上面的代码。它工作正常,也得到了正确的结果!
有关parallel_reduce的更多信息,请参阅以下链接:
https://software.intel.com/content/www/us/en/develop/documentation/tbb-documentation/top/intel-threading-building-blocks-developer-reference/algorithms/parallelreduce-template-function.html
https://link.springer.com/content/pdf/10.1007%2F978-1-4842-4398-5.pdf
谢谢,
桑托什
我想调用parallel_reduce对向量元素求和。但是我发现如果向量元素足够了,结果是不正确的。请帮助我如何使用此功能。
// prepare data
const size_t allNum = 1000000;
std::vector<double> a;
for (int i = 0; i < allNum; ++i)
{
a.push_back(double(i + 1));
}
// λ func
auto f = [&]() -> double {
return tbb::parallel_reduce(tbb::blocked_range<size_t>(0, allNum),
0.0,
[&](const tbb::blocked_range<size_t>& r, double init) -> double {
for (int i = r.begin(); i < r.end(); ++i)
{
init += a[i];
}
return init;
},
[](double f, double s) -> double {
return f + s;
}
/*std::plus<double>()*/);
};
// call λ func, get the result
double correctResult = (1.0 + 1000000.0) * 500000.0;
double sum = f(); // sum != correctResult
// sum is different every loop
我试过运行上面的代码。它工作正常,也得到了正确的结果!
有关parallel_reduce的更多信息,请参阅以下链接: https://software.intel.com/content/www/us/en/develop/documentation/tbb-documentation/top/intel-threading-building-blocks-developer-reference/algorithms/parallelreduce-template-function.html
https://link.springer.com/content/pdf/10.1007%2F978-1-4842-4398-5.pdf
谢谢, 桑托什