Linux求数组长度为10000的浮点数(精确小数点右4位)计算值
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发布时间：2022-04-27 09:20
共1个回答
热心网友
时间：2023-10-03 05:55
既然题主没有说要求用什么语言，那我就用c++11实现了。
#include <iostream>
#include <random>
#include <thread>
#include <chrono>
#include <algorithm>
#include <iomanip>
using namespace std;
const int size = 10000;
float table[size];
int main(){
random_device engine;
uniform_real_distribution<float> dist(0, 1);
float sum;
for(auto& i: table){
i = dist(engine);
}
auto t_start = chrono::system_clock::now();
sum = accumulate(table, table + size, 0.0);
auto t_end = chrono::system_clock::now();
auto ration = std::chrono::ration_cast<std::chrono::microseconds>(t_end - t_start).count();
cout << "sum of the main thread: " << fixed << setprecision(4) << sum << endl;
cout << "time elapsed: " << ration << " micro seconds" << endl;
float sum_child[4];
auto fun = [&](int index){
sum_child[index] = accumulate(table + index * size / 4, table + (index + 1) * size / 4, 0.0);
};
t_start = chrono::system_clock::now();
thread thrd_table[4] = {
thread(fun, 0), thread(fun, 1), thread(fun, 2), thread(fun, 3)
};
for(auto& thrd: thrd_table){
thrd.join();
}
sum = 0;
sum = accumulate(sum_child, sum_child + 4, 0.0);
t_end = chrono::system_clock::now();
ration = std::chrono::ration_cast<std::chrono::microseconds>(t_end - t_start).count();
cout << "sum of child threads: " << fixed << setprecision(4) << sum << endl;
cout << "time elapsed: " << ration << " micro seconds" << endl;
return 0;
}
编译：
g++ -std=c++11 test.cc -lpthread -o test
运行：
./test
结果：
sum of the main thread: 4976.8721
time elapsed: 0 ms
sum of child threads: 4976.8721
time elapsed: 0 ms
由于随机性每次加和的数值不同，但是精确到毫秒时，时间测出来妥妥的都是零。就是数据量太小，实际运行时间在微秒量级，当然看不出来。
精度改为微秒以后：
sum of the main thread: 4957.9878
time elapsed: 113 micro seconds
sum of child threads: 4957.9878
time elapsed: 560 micro seconds
多线程反而比单线程慢，因为启动线程本身也需要时间。
数据量再增大1000倍：
sum of the main thread: 4999892.0000
time elapsed: 25313 micro seconds
sum of child threads: 4999892.0000
time elapsed: 8986 micro seconds
这回看着正常多了吧
热心网友
时间：2023-10-03 05:55
既然题主没有说要求用什么语言，那我就用c++11实现了。
#include <iostream>
#include <random>
#include <thread>
#include <chrono>
#include <algorithm>
#include <iomanip>
using namespace std;
const int size = 10000;
float table[size];
int main(){
random_device engine;
uniform_real_distribution<float> dist(0, 1);
float sum;
for(auto& i: table){
i = dist(engine);
}
auto t_start = chrono::system_clock::now();
sum = accumulate(table, table + size, 0.0);
auto t_end = chrono::system_clock::now();
auto ration = std::chrono::ration_cast<std::chrono::microseconds>(t_end - t_start).count();
cout << "sum of the main thread: " << fixed << setprecision(4) << sum << endl;
cout << "time elapsed: " << ration << " micro seconds" << endl;
float sum_child[4];
auto fun = [&](int index){
sum_child[index] = accumulate(table + index * size / 4, table + (index + 1) * size / 4, 0.0);
};
t_start = chrono::system_clock::now();
thread thrd_table[4] = {
thread(fun, 0), thread(fun, 1), thread(fun, 2), thread(fun, 3)
};
for(auto& thrd: thrd_table){
thrd.join();
}
sum = 0;
sum = accumulate(sum_child, sum_child + 4, 0.0);
t_end = chrono::system_clock::now();
ration = std::chrono::ration_cast<std::chrono::microseconds>(t_end - t_start).count();
cout << "sum of child threads: " << fixed << setprecision(4) << sum << endl;
cout << "time elapsed: " << ration << " micro seconds" << endl;
return 0;
}
编译：
g++ -std=c++11 test.cc -lpthread -o test
运行：
./test
结果：
sum of the main thread: 4976.8721
time elapsed: 0 ms
sum of child threads: 4976.8721
time elapsed: 0 ms
由于随机性每次加和的数值不同，但是精确到毫秒时，时间测出来妥妥的都是零。就是数据量太小，实际运行时间在微秒量级，当然看不出来。
精度改为微秒以后：
sum of the main thread: 4957.9878
time elapsed: 113 micro seconds
sum of child threads: 4957.9878
time elapsed: 560 micro seconds
多线程反而比单线程慢，因为启动线程本身也需要时间。
数据量再增大1000倍：
sum of the main thread: 4999892.0000
time elapsed: 25313 micro seconds
sum of child threads: 4999892.0000
time elapsed: 8986 micro seconds
这回看着正常多了吧
