Question

我成功地编写了一个程序，以给定三个系数找到标准形式二次方程的解。我现在正在尝试简化结果，以消除底部的“ / 2a”。因此是simple（）函数。

我已经开始创建简化函数，并试图将解的虚部和实部除以2a。不知何故，它给出了错误：

"error:invalid operands of types 'int' and 'double *' to binary 'operator*'" on lines 105 and 106.

我怀疑这与我作为参数传递的指针和引用有关。我是这个主意。

此外，仅一句，我从未真正看到过使用“ / =”。是允许的吗？我知道“ + =”可以。

#include <iostream>
#include <string>
#include <cmath>
#include <cstdlib> //simplify the answer
using namespace std;

int count = 0;
//prototyping
double ans_1(double, double, double);
double ans_2(double, double, double);
double nanhe(double, double, double);
void simplify(double*, double*, double*);

int main()
{
    double a, b, c;
    cout << "Quadratic Equation Solver \n";
    cout << "Enter a value for a: ";
    cin >> a;
    cout << endl;
    cout << "Enter a value for b: ";
    cin >> b;
    cout << endl;
    cout << "Enter a value for c: ";
    cin >> c;
    cout << endl;

    if (isnan(sqrt((b * b) - (4 * a * c)))) {
        count++;
    }
    if (!count) {
        double answer01 = ans_1(a, b, c);
        double answer02 = ans_2(a, b, c);
        cout << "X=" << answer01 << endl;
        cout << "X=" << answer02 << endl;
    }
    else if (count) //could route these imag ones to separate funcitons instead of count++
    {
        double answer01 = ans_1(a, b, c);
        double answer02 = ans_1(a, b, c);
        cout << "X=(" << -b << "+" << answer01 << "i)/" << 2 * a << endl;
        cout << "X=(" << -b << '-' << answer02 << "i)/" << 2 * a << endl;
    }
    system("pause");
}

double ans_1(double a, double b, double c)
{
    double ans1;
    double temp_c = sqrt((b * b) - (4 * a * c));
    if (isnan(temp_c)) {
        temp_c = nanhe(a, b, c);
    }
    if (!count) {
        ans1 = ((-b + temp_c) / (2 * a));
    }
    else if (count) {
        ans1 = ((temp_c));
    }
    simplify(&a, &b, &ans1);
    return ans1;
}

double ans_2(double a, double b, double c)
{
    double ans2;
    double temp_d = sqrt((b * b) - (4 * a * c));
    if (isnan(temp_d)) {
        temp_d = nanhe(a, b, c);
    }
    if (!count) {
        ans2 = ((-b - temp_d) / (2 * a));
    }
    else if (count) {
        ans2 = (temp_d);
    }
    simplify(&a, &b, &ans2); //line under this should alter ans2 so its returning the simplified version instead, or just make a new variable instead of ans2
    return ans2;
}

double nanhe(double a, double b, double c) //still need to apply simplify() to nanhe
{
    double temp_help;
    temp_help = sqrt(-1 * ((b * b) - (4 * a * c)));
    count++;
    return temp_help;
}

void simplify(double* a, double* b, double* ans) //only run if complex
{
    ans /= (2 * a);
    b /= (2 * a);
}

Answer 1

您需要访问或更改指针指向的值，而不是指针本身：

void simplify(double* a, double* b, double* ans) //only run if complex
{
    *ans /= (2 * *a);
    *b /= (2 * *a);
}

编辑： 或者，如@Hans Passant所说：

并考虑到C ++允许您将其参数声明为double＆而不是double *。并去购买像样的IDE，这样您就不必猜测语法错误的位置。

使用指针的二次方程问题

1 个答案: