C++ 取消引用具有相同地址的指针会返回不同的结果

这是我的密码：

#include "stdafx.h"
#include "math.h"
#include <iostream>

using namespace std;

double Calc_H(double Q, double Head, double *constants)
{
    return (constants[0] * pow(Q, 4) + constants[1] * pow(Q, 3) + constants[2] * pow(Q, 2) + constants[3] * Q + constants[4] - Head);
}

double Calc_dH(double Q, double *constants)
{
    return (4 * constants[0] * pow(Q, 3) + 3 * constants[1] * pow(Q, 2) + 2 * constants[2] * Q + constants[3]);
}

double NewtonRaphson(double Head, double first_guess, double max_error, double * constants)
{
    double Q_iter = first_guess;
    int iter_counter = 1;
    cout << constants << endl << constants[0] << endl << constants[1] << endl;
    while (abs(Calc_H(Q_iter, Head, constants)) > max_error || iter_counter > 1000)
    {
        Q_iter = Q_iter - Calc_H(Q_iter, Head, constants) / Calc_dH(Q_iter, constants);
        iter_counter++;
    }
    return Q_iter;
}

double * Calc_constants(double freq)
{
    double * pointer;
    double constants[6];
    constants[0] = -1.2363 + 2.3490 / 10 * freq - 1.3754 / 100 * pow(freq, 2) + 2.9027 / 10000 * pow(freq, 3) - 2.0004 / 1000000 * pow(freq, 4);
    constants[1] = 1.9547 - 4.5413 / 10 * freq + 3.5392 / 100 * pow(freq, 2) - 8.1716 / 10000 * pow(freq, 3) + 5.9227 / 1000000 * pow(freq, 4);
    constants[2] = -5.3522 - 4.5413 / 10 * freq - 1.3311 / 100 * pow(freq, 2) + 4.8787 / 10000 * pow(freq, 3) - 4.8767 / 1000000 * pow(freq, 4);
    constants[3] =  3.8894 / 100 + 3.5888 / 10 * freq + 1.0024 / 100 * pow(freq, 2) - 5.6565 / 10000 * pow(freq, 3) + 7.5172 / 1000000 * pow(freq, 4);
    constants[4] = -8.1649 + 5.4525 / 10 * freq - 3.2415 / 100 * pow(freq, 2) + 8.9033 / 10000 * pow(freq, 3) - 9.0927 / 1000000 * pow(freq, 4);
    constants[5] =  2.1180 / 10 + 5.0018 / 100 * freq + 6.0490 / 1000 * pow(freq, 2) - 1.5707 / 100000 * pow(freq, 3) + 3.7572 / 10000000 * pow(freq, 4);

    pointer = constants;
    return pointer;
}

int _tmain(int argc, _TCHAR* argv[])
{

    double * constants;
    //Determine constants based on freq (see manual pump)
    double freq;
    cin >> freq; 
    double head;
    cin >> head;
    constants = Calc_constants(freq);
    cout << constants[0] << endl << constants[1] << endl << constants << endl;
    cout << NewtonRaphson(head, 0, 0.001, constants) << endl;
    cin >> freq;    
    return 0;
}

Calc_常量在堆栈（而不是堆）上为此数组分配内存。 当此函数返回时，此内存块可能被分配用于其他目的，因此不应在此函数之外访问。 因此，当指针返回并在以后使用时，会导致不可预知的结果

常量数组需要在main或heap中分配，所以它的生存期对于这种使用来说足够长

在这个while循环条件下

while (abs(Calc_H(Q_iter, Head, constants)) > max_error || iter_counter > 1000)

---

我猜，应该是ItdioTrave< 1000。

< P>代码中没有太多C++。首先，让我们删除非标准内容：

---

然后到处都是C风格的数组。你不想那样做。使用

std:：array

或

std:：vector

，问题将自行消失

下面是一个使用

std:：vector

的示例：

#include <math.h>
#include <iostream>
#include <vector>

double Calc_H(double Q, double Head, std::vector<double> const& constants)
{
    return (constants[0] * pow(Q, 4) + constants[1] * pow(Q, 3) + constants[2] * pow(Q, 2) + constants[3] * Q + constants[4] - Head);
}

double Calc_dH(double Q, std::vector<double> const& constants)
{
    return (4 * constants[0] * pow(Q, 3) + 3 * constants[1] * pow(Q, 2) + 2 * constants[2] * Q + constants[3]);
}

double NewtonRaphson(double Head, double first_guess, double max_error, std::vector<double> const& constants)
{
    double Q_iter = first_guess;
    int iter_counter = 1;
    std::cout << constants.data() << std::endl << constants[0] << std::endl << constants[1] << std::endl;
    while (abs(Calc_H(Q_iter, Head, constants)) > max_error && iter_counter < 1000)
    {
        Q_iter = Q_iter - Calc_H(Q_iter, Head, constants) / Calc_dH(Q_iter, constants);
        iter_counter++;
    }
    return Q_iter;
}

std::vector<double> Calc_constants(double freq)
{
    std::vector<double> constants(6);
    constants[0] = -1.2363 + 2.3490 / 10 * freq - 1.3754 / 100 * pow(freq, 2) + 2.9027 / 10000 * pow(freq, 3) - 2.0004 / 1000000 * pow(freq, 4);
    constants[1] = 1.9547 - 4.5413 / 10 * freq + 3.5392 / 100 * pow(freq, 2) - 8.1716 / 10000 * pow(freq, 3) + 5.9227 / 1000000 * pow(freq, 4);
    constants[2] = -5.3522 - 4.5413 / 10 * freq - 1.3311 / 100 * pow(freq, 2) + 4.8787 / 10000 * pow(freq, 3) - 4.8767 / 1000000 * pow(freq, 4);
    constants[3] =  3.8894 / 100 + 3.5888 / 10 * freq + 1.0024 / 100 * pow(freq, 2) - 5.6565 / 10000 * pow(freq, 3) + 7.5172 / 1000000 * pow(freq, 4);
    constants[4] = -8.1649 + 5.4525 / 10 * freq - 3.2415 / 100 * pow(freq, 2) + 8.9033 / 10000 * pow(freq, 3) - 9.0927 / 1000000 * pow(freq, 4);
    constants[5] =  2.1180 / 10 + 5.0018 / 100 * freq + 6.0490 / 1000 * pow(freq, 2) - 1.5707 / 100000 * pow(freq, 3) + 3.7572 / 10000000 * pow(freq, 4);

    return constants;
}

int main()
{
    //Determine constants based on freq (see manual pump)
    double freq;
    std::cin >> freq; 
    double head;
    std::cin >> head;
    std::vector<double> constants = Calc_constants(freq);
    std::cout << constants[0] << std::endl << constants[1] << std::endl << constants.data() << std::endl;
    std::cout << NewtonRaphson(head, 0, 0.001, constants) << std::endl;
    std::cin >> freq;    
    return 0;
}

这只会产生未定义的行为<代码>常量是一个局部变量。当函数返回时，您在这里创建的六个元素将被销毁，但您仍保留一个指向它们的指针。C++语言不能保证会发生什么，如果你试图在以后引用那个指针（就像你那样）。如果运气好的话，程序会立即崩溃，向您显示存在严重错误，而不是生成无意义的输出

你也有点不走运，没有得到编译器的警告。如果您没有使用冗余的

指针

变量，那么您可能会收到警告（至少在VC 2013中是这样）

简单的例子：

double * Calc_constants()
{
    double constants[6];
    return constants;
}

int main()
{
    double* ptr = Calc_constants();
}

VC 2013警告：

warning C4172: returning address of local variable or temporary

---

使用

std:：vector

，数据在内部分配，以便您可以安全地返回对象。您可以像使用简单的

int

s一样安全地使用标准容器对象，而不会将原始指针的复杂性撒在代码上。

谢谢，我会尝试解决这个问题。。。。再次感谢您纠正了另一个错误：）可能重复的

int main()

#include <math.h>
#include <iostream>
#include <vector>

double Calc_H(double Q, double Head, std::vector<double> const& constants)
{
    return (constants[0] * pow(Q, 4) + constants[1] * pow(Q, 3) + constants[2] * pow(Q, 2) + constants[3] * Q + constants[4] - Head);
}

double Calc_dH(double Q, std::vector<double> const& constants)
{
    return (4 * constants[0] * pow(Q, 3) + 3 * constants[1] * pow(Q, 2) + 2 * constants[2] * Q + constants[3]);
}

double NewtonRaphson(double Head, double first_guess, double max_error, std::vector<double> const& constants)
{
    double Q_iter = first_guess;
    int iter_counter = 1;
    std::cout << constants.data() << std::endl << constants[0] << std::endl << constants[1] << std::endl;
    while (abs(Calc_H(Q_iter, Head, constants)) > max_error && iter_counter < 1000)
    {
        Q_iter = Q_iter - Calc_H(Q_iter, Head, constants) / Calc_dH(Q_iter, constants);
        iter_counter++;
    }
    return Q_iter;
}

std::vector<double> Calc_constants(double freq)
{
    std::vector<double> constants(6);
    constants[0] = -1.2363 + 2.3490 / 10 * freq - 1.3754 / 100 * pow(freq, 2) + 2.9027 / 10000 * pow(freq, 3) - 2.0004 / 1000000 * pow(freq, 4);
    constants[1] = 1.9547 - 4.5413 / 10 * freq + 3.5392 / 100 * pow(freq, 2) - 8.1716 / 10000 * pow(freq, 3) + 5.9227 / 1000000 * pow(freq, 4);
    constants[2] = -5.3522 - 4.5413 / 10 * freq - 1.3311 / 100 * pow(freq, 2) + 4.8787 / 10000 * pow(freq, 3) - 4.8767 / 1000000 * pow(freq, 4);
    constants[3] =  3.8894 / 100 + 3.5888 / 10 * freq + 1.0024 / 100 * pow(freq, 2) - 5.6565 / 10000 * pow(freq, 3) + 7.5172 / 1000000 * pow(freq, 4);
    constants[4] = -8.1649 + 5.4525 / 10 * freq - 3.2415 / 100 * pow(freq, 2) + 8.9033 / 10000 * pow(freq, 3) - 9.0927 / 1000000 * pow(freq, 4);
    constants[5] =  2.1180 / 10 + 5.0018 / 100 * freq + 6.0490 / 1000 * pow(freq, 2) - 1.5707 / 100000 * pow(freq, 3) + 3.7572 / 10000000 * pow(freq, 4);

    return constants;
}

int main()
{
    //Determine constants based on freq (see manual pump)
    double freq;
    std::cin >> freq; 
    double head;
    std::cin >> head;
    std::vector<double> constants = Calc_constants(freq);
    std::cout << constants[0] << std::endl << constants[1] << std::endl << constants.data() << std::endl;
    std::cout << NewtonRaphson(head, 0, 0.001, constants) << std::endl;
    std::cin >> freq;    
    return 0;
}

double * Calc_constants(double freq)
{
    double * pointer;
    double constants[6];
    // ...    
    pointer = constants;
    return pointer;
}

double * Calc_constants()
{
    double constants[6];
    return constants;
}

int main()
{
    double* ptr = Calc_constants();
}

warning C4172: returning address of local variable or temporary