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Use long double for check solution in case of very large problem sizes

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Tom Deakin 2022-04-01 10:51:24 +01:00
parent e77a34158c
commit 6185d3aca6
2 changed files with 6 additions and 5 deletions
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1
CHANGELOG.md
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@ -3,6 +3,7 @@ All notable changes to this project will be documented in this file.
## Unreleased
- Fix CUDA memory limit check.
- Use long double for `check_solution` in case of large problem size.
## [v4.0] - 2021-12-22

10
src/main.cpp
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@ -487,15 +487,15 @@ void check_solution(const unsigned int ntimes, std::vector<T>& a, std::vector<T>
goldSum = goldA * goldB * ARRAY_SIZE;
// Calculate the average error
double errA = std::accumulate(a.begin(), a.end(), 0.0, [&](double sum, const T val){ return sum + fabs(val - goldA); });
long double errA = std::accumulate(a.begin(), a.end(), 0.0, [&](double sum, const T val){ return sum + fabs(val - goldA); });
errA /= a.size();
double errB = std::accumulate(b.begin(), b.end(), 0.0, [&](double sum, const T val){ return sum + fabs(val - goldB); });
long double errB = std::accumulate(b.begin(), b.end(), 0.0, [&](double sum, const T val){ return sum + fabs(val - goldB); });
errB /= b.size();
double errC = std::accumulate(c.begin(), c.end(), 0.0, [&](double sum, const T val){ return sum + fabs(val - goldC); });
long double errC = std::accumulate(c.begin(), c.end(), 0.0, [&](double sum, const T val){ return sum + fabs(val - goldC); });
errC /= c.size();
double errSum = fabs((sum - goldSum)/goldSum);
long double errSum = fabs((sum - goldSum)/goldSum);
double epsi = std::numeric_limits<T>::epsilon() * 100.0;
long double epsi = std::numeric_limits<T>::epsilon() * 100.0;
if (errA > epsi)
std::cerr