函数调用错误

Function Call Error

本文关键字：错误 函数调用      更新时间：2023-10-16

到目前为止，除了函数 Pelly 之外，我的代码都正常工作。它不会像预期的那样返回调整后的毛额。 我非常确定数学是正确的，我认为问题在于函数的调用方式。我不太擅长功能。任何帮助将不胜感激。

#include <iostream>
using namespace std;
int main()
{
    double Federal = 0, PellGrant = 5730, AdjustedGross = 0, Total = 0;
    int YesNo;
    int const StaffordLoan = 9500;
    cout << "Let me forecast your FAFSA" << endl;
    cout << "Enter your adjusted Gross income: " << endl; cin >> AdjustedGross;
    if (AdjustedGross >= 30000)
    {
        cout << "Sorry, your income is too high for this forecaster";
        return 0;
    }
    cout << "Can someone claim you as a dependent? [1 = yes / 0 = no]: " << endl; cin >> YesNo;
    if (YesNo == 1)
    {
        PellGrant -= 750;
    }
    Federal = 1465;
    if (AdjustedGross >= 19000)
    {
        cout << "I'm sorry, but the Work-Study Award is not available to you" << endl;
        Federal = 0;
    }
    double Pelly(AdjustedGross);
    Total = Federal + StaffordLoan + PellGrant;
    if (Federal != 0)
    {
        cout << "Your Work-Study Award (if available): " << Federal << endl;
    }
    cout << "Your Stafford Loan award (if needed): " << StaffordLoan << endl;
    cout << "Your Pell Grant: " << PellGrant << endl;
    return (0);
}
double  Pelly(double x)
{
    // x is AdjustedGross
    if ((x > 12000) && (x < 20000)) // make sure adjusted gross is bettween 12000 & 20000
    {
        double a = x / 1000; // for every 1000 in adjusted, subtract 400
            a *= 400;
        x -= a;
    }
    if (x > 20000) // check adjusted > 20000
    {
        double  a = x / 1000; // for every 1000 in adjusted, subtract 500
        a *= 500;
        x -= a;
    }
    return x;
}

制作：

double Pelly(AdjustedGross);

到：

double foo = Pelly(AdjustedGross);

将从Pelly返回的值存储在double变量foo中。在函数Pelly上使用前向声明，换句话说，在main之前这样声明它：

double Pelly(double);

您需要实际将函数的结果分配给变量，然后使用此结果。所以你应该做这样的事情：

double pellyResult = Pelly(AdjustedGross);

您还应该确保在 main 上方声明您的函数：

double pellyResult(double);

方法的签名应该是

void Pelly(double& AdjustedGross)

即根本没有返回值(这样，AdjustedGross 通过引用传递并直接在函数内部修改，然后调用函数将是

Pelly(AdjustedGross);

或者你的函数调用应该是 

double foo = Pelly(AdjustedGross)

如其他答案所述。

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