Question

我正在尝试制作一个细菌模型只是为了好玩，我正在使用pow（a，b）函数作为计算人口增长的函数。当细菌数量达到其环境所能提供的最大食物单位数量时，由于个体之间的竞争，它会下降70％。我将结果保存为txt，以便稍后绘制。我遇到的问题是群体正确振荡，直到我达到t = 900的多个再现周期，然后群体默认为0。代码如下，我希望你不介意用葡萄牙语写的变量和函数的名称。

bool
check_aliemento (unsigned long int *pop)
{
    if (*pop >= MAX_ALIMENTO) return false;
    return true;
}

unsigned long int
replicaBacteria (unsigned long int *popInit, unsigned int tempo_t, double taxa)
{
    unsigned long int nextPop = round ((*popInit) * 
                static_cast<double> (pow (1 + taxa, tempo_t))); 
                //I'm almost sure that the problem happens in this pow() function
    while (! check_aliemento (&nextPop))
    {
        nextPop = (0.7 * nextPop);
    }
    return nextPop;
}



   int
main ( int argc, char** argv )
{
    unsigned long int a = 2;
    ofstream myfile;
    myfile.open ( "C:\\Users\\Pedro\\Desktop\\values.txt" );
    for ( unsigned int i; i < 1000; i ++ )
    {
        unsigned long int pop = replicaBacteria ( &a, i, 0.05 );
        myfile << pop << " ==> time = " << i;
        myfile << "\r\n";
    }
    myfile.close ( );
    return 0;
}

示例输出：

8080 ==> time = 872
8484 ==> time = 873
8909 ==> time = 874
9354 ==> time = 875
9822 ==> time = 876
7219 ==> time = 877
7580 ==> time = 878
7958 ==> time = 879
8357 ==> time = 880
8775 ==> time = 881
9214 ==> time = 882
9675 ==> time = 883
7110 ==> time = 884
7466 ==> time = 885
7839 ==> time = 886
8232 ==> time = 887
8643 ==> time = 888
9075 ==> time = 889
9529 ==> time = 890
7003 ==> time = 891
7354 ==> time = 892
7721 ==> time = 893
8108 ==> time = 894
8513 ==> time = 895
0 ==> time = 896
0 ==> time = 897
0 ==> time = 898
0 ==> time = 899
0 ==> time = 900

Answer 1

您正在溢出long int nextPop。这是因为你正在计算人口增长，好像它是896代不间断的，并且只有在那之后减少人口直到它低于你的魔法限制（这里是10,000，你说）。

解决这个问题的一种方法是在每次迭代后保留数量，并将其增长5％，然后在必要时将结果减少到70％。

所以代替unsigned long int pop = replicaBacteria ( &a, i, 0.05 );，你需要像pop = replicaBacteria ( pop, 0.05 );这样的东西（在循环之前声明变量），然后replicaBacteria应该简单地将其输入乘以5％并缩放到70％必要时的结果。

编辑代码以生成可编译并运行的内容

像这样：

#include <cmath>
#include <fstream>
#include <iostream>

#define MAX_ALIMENTO 10000

double
replicaBacteria (double popInit, double taxa)
{
    double nextPop = popInit * (1 + taxa); 
    while (nextPop >= MAX_ALIMENTO)
    {
        nextPop = (0.7 * nextPop);
    }
    return nextPop;
}



   int
main ( int argc, char** argv )
{
    std::ofstream myfile( "/tmp/out.txt" );
    double pop = 2;
    unsigned int i;
    for ( i=0; i < 1000; i ++ )
    {
        pop = replicaBacteria ( pop, 0.05 );
        myfile << round(pop) << " ==> time = " << i;
        myfile << "\r\n";
    }
    myfile.close ( );
    return 0;
}

输出：

2 ==> time = 0
2 ==> time = 1
2 ==> time = 2
2 ==> time = 3
3 ==> time = 4
3 ==> time = 5
3 ==> time = 6
3 ==> time = 7
3 ==> time = 8
3 ==> time = 9
3 ==> time = 10
4 ==> time = 11
...
8925 ==> time = 990
9371 ==> time = 991
9840 ==> time = 992
7232 ==> time = 993
7594 ==> time = 994
7974 ==> time = 995
8372 ==> time = 996
8791 ==> time = 997
9230 ==> time = 998
9692 ==> time = 999

Answer 2

pow (1 + taxa, tempo_t)

显然long int无法保留2^900所以你在这里看到的是整数溢出使用的数据类型可以包含大值，如

unsigned long long

Answer 3

甚至unsigned long long int也无法成功。当tempo_t＆gt; 895然后

unsigned long long int nextPop = round ((*popInit) * 
                    static_cast<double> (pow (1 + taxa, tempo_t)));

此表达式评估为零，因此您必须缩短细菌的使用寿命。

Answer 4

稍微改变你的功能，以了解数字何时变为零：

unsigned long int
replicaBacteria (unsigned long int *popInit, unsigned int tempo_t, double taxa)
{
   double r = round ((*popInit) * static_cast<double> (pow (1 + taxa, tempo_t))); 
   std::cout << r << std::endl;
   unsigned long int nextPop = r;
   std::cout << nextPop << std::endl;

   //I'm almost sure that the problem happens in this pow() function
   while (! check_aliemento (&nextPop))
   {
      nextPop = (0.7 * nextPop);
   }
   return nextPop;
}

我看到以下输出：

1.84269e+19
18426916303946758144
8513 ==> time = 895
1.93483e+19
0
0 ==> time = 896
2.03157e+19
0
0 ==> time = 897

这是在使用g ++ 4.8.2的64位Linux机器上。由于从896开始看到零，因此可以安全地假设您的编译器具有完全相同的问题。 unsigned long int可以表示的最大数字小于1.93483e+19。

人口增长模型与预期不符

4 个答案: