相关疑难解决方法(0)

我想我有一个非常糟糕的概念问题.为什么我简单地用valgrind获得了很多竞争条件错误.首先我认为这可能是一个错误,我在论坛中看到linux的更新滚动版本将解决这个问题...所以现在我已经打开了tubleweed,100%更新.以下代码muss有一些非常错误:

#include <iostream>
#include <thread>

using namespace std;

class FOO
{
public:
    void do_something ()
    {
        cout<<"cout somethin\n";
    }
};

int main ()
{
    FOO foo;
    std::thread t (&FOO::do_something,&foo);
    t.join();
}

当我做的时候

valgrind --tool=drd ./oncordia

我必须尝试5次或更多,我得到以下错误:

==6218== drd, a thread error detector
==6218== Copyright (C) 2006-2010, and GNU GPL'd, by Bart Van Assche.
==6218== Using Valgrind-3.6.1 and LibVEX; rerun with -h for copyright info
==6218== Command: ./oncordia
==6218== 
cout somethin
==6218== Conflicting store by thread 1 at 0x05b5d050 size 8 …

这是一个算法问题.为了简单起见,我说我有两个双打,A和B.我想构建一个函数,它会给我差异,直到A的下一个倍数或B的下一个倍数,如果这有意义的话.

例如,假设A是3而B是5.

考虑倍数:(3,6,9,12,15)和(5,10,15).

我希望函数输出:(3,2,1,3,1,2,3),因为它需要3个单位才能达到3,然后再多达2个到达5,然后是1到6,然后是3到9等...

我希望这是有道理的.理想情况下,它是一个Python-esque生成器(虽然我在Arduino~C++中编写它).我需要快速 - 非常快.

真的很感激任何帮助.我的伪代码在下面,但它不是那么好.

a = 3
b = 5

current = 0
distToA = a
distToB = b
for i in xrange(100):
  if distToA > distToB: #B comes first
    print "Adding {0}".format(distToB)
    current += distToB
    distToA -= distToBb
    distToB = b
  elif distToB > distToA: #A comes first
    print "Adding {0}".format(distToA)
    current += distToA
    distToB -= distToA
    distToA = a
  else: #Equal
    print "Adding {0}".format(distToA)
    current += distToA #Arbitrarily, could be distToB
    distToA = a …