Question

编辑：全局变量作为作业的一部分受到限制（忘了提及）。

该计划旨在计算指定年数的每年预计人口。这是等式：

N = P(1 + B)(1 - D)

其中N是新的人口规模，P是先前的人口规模，B是出生率，D是死亡率。 B和D是小数。

所以我的问题是我只能获得第一次迭代的结果，而我不知道如何更新结果以便它乘以（1 + B）和（1-D）。我的猜测是它与其中一个组合赋值运算符有关，但我没有成功。

#include <iostream>
#include <iomanip>
#include <cmath>

using namespace std;

int get_pop();
int get_numYears();
double get_annBirthRate();
double get_annDeathRate();
double pop_new(int& A, double& B, double& C, double& D);

int main()
{
    int P,
    num,
    Y;

    double B,
    D,
    N;

    P = get_pop();
    B = get_annBirthRate();
    D = get_annDeathRate();
    Y = get_numYears();
    for (num = 1; num != Y; num++)
    {
        N = pop_new(P, B, D, N);
        cout << "\n\nThe projected population for the end of year " << num << " is " << N << endl;
        cin.ignore();
    }

    return 0;
}

int get_pop()
{
    int pop;

    cout << "Enter the starting population:" << endl;
    cin  >> pop;
    while (pop < 2)
    {
        cout << "\n\nError - starting population cannot be less than 2:" << endl;
        cin  >> pop;
    }
    return pop;
}


double get_annBirthRate()
{
    double annBirthRate;

    cout << "\n\nEnter the annual birth rate \n(a positive percentage in decimal form):" << endl;
    cin  >> annBirthRate;
    while (annBirthRate < 0)
    {
        cout << "\n\nError - annual birth rate cannot be negative:" << endl;
        cin  >> annBirthRate;
    }
    return annBirthRate;
}


double get_annDeathRate()
{
    double annDeathRate;

    cout << "\n\nEnter the annual death rate \n(a positive percentage in decimal form):" << endl;
    cin  >> annDeathRate;
    while (annDeathRate < 0)
    {
        cout << "\n\nError - death rate cannot be negative:" << endl;
        cin  >> annDeathRate;
    }
    return annDeathRate;
}


int get_numYears()
{
    int numYears;

    cout << "\n\nEnter the number of years to display:" << endl;
    cin  >> numYears;
    while (numYears < 0)
    {
        cout << "\n\nError - number of years cannot be less than 1:" << endl;
        cin  >> numYears;
    }
    return numYears;
}

double pop_new(int& P, double& B, double& D, double& N)
{
    N = P * (1 + B) * (1 - D);

    return N;
}

Answer 1

值P，B和D不会改变。

成功：

double pop_new(int P, double B, double D)
{
   double N = P * (1 + B) * (1 - D);
   return N;
}

和

P = pop_new(P, B, D);
cout << "\n\nThe projected population for the end of year " 
     << num << " is " << P << endl;

注意：你不需要传递引用（如果你应该传递const double＆amp;）

更新迭代结果

1 个答案: