'('用于函数式转换或构造类型Xcode错误

时间:2015-11-17 01:57:51

标签: c++ xcode function parameters void

This is where I get error:
 calcScore(double score1,double score2, double score3, double score4, double score5);

这是我的代码,我不知道如何修复错误;我不确定错误是什么意思。我试过看教科书和网上的例子,但我仍然感到困惑;这是我们必须使用函数

的第一个任务
#include <iostream>
using namespace std;

void getJudgesData();
void calcScore();

int main()
{
    double scoreAverage=-9;
    double score1=-9.0;
    double score2=-9.0;
    double score3=-9.0;
    double score4=-9.0;
    double score5=-9.0;

    getJudgesData();

    calcScore(double score1,double score2, double score3, double score4, double score5);

    return 0;
}

 // getJudgesData Program
void getJudgesData(double &score1, double &score2, double &score3, double &score4, double &score5)
{

    do
    {
        cout << "Enter judges score: ";
        cin >> score1;
        do
        {
            if (score1 < 0 || score1 >10)
            {
                cout << "Score has to range from 0-10 \n";
                cout << "Please enter a valid score: ";
                cin >> score1;
            }
        }while (score1<0 || score1 >10);
   }
    while (score1<0 || score1 >10);

    do
    {
        cout << "Enter judges score: ";
        cin >> score2;
        do
        {
            if (score2 < 0 || score2 >10)
            {
                cout << "Score has to range from 0-10 \n";
                cout<< "Please enter a valid score: ";
                cin >> score2;
            }
        }while (score2 <0 || score2 >10);
    }
    while (score2<0 || score2 >10);

    do
    {
        cout << "Enter judges score: ";
        cin >> score3;
        do
        {
            if (score3 < 0 || score3 >10)
           {
                cout << "Score has to range from 0-10 \n";
                cout << "Please enter a valid score: ";
                cin >> score3;
            }
        }while (score3 < 0 || score3 >10);
    }
    while (score3<0 || score3 >10);

    do
    {
        cout << "Enter judges score: ";
        cin >> score4;
        do
        {
            if (score4 < 0 || score4 >10)
            {
                cout << "Score has to range from 0-10 \n";
                cout << "Please enter a valid score: ";
                cin >> score4;
            }
        }while (score4 < 0 || score4 >10);
    }
    while (score4<0 || score4 >10);

    do
    {
        cout << "Enter judges score: ";
        cin >> score5;
        do
        {
            if (score5 < 0 || score5 >10)
            {
                cout << "Score has to range from 0-10 \n";
                cout << "Please enter a valid score: ";
                cin >> score5;
            }
        }while (score5 < 0 || score5 >10);
    }
    while (score5<0 || score5 >10);

    return;
}

//Start of calcScore
void calcScore( double score1, double score2, double score3, double score4,double score5)
{
    double average;

    average = (score1+score2+score3+score4+score5)/ 5;

    return;
}

1 个答案:

答案 0 :(得分:0)

让我们先处理你的错误:正如@ shree.pat18在评论中指出的那样,你正在调用一个函数并尝试给出参数的类型。这是不允许的。 C ++ 知道参数的类型,因为您之前已经声明了它们。您可以这样致电calcScore

double score1=-9.0;
double score2=-9.0;
double score3=-9.0;
double score4=-9.0;
double score5=-9.0;

getJudgesData();

calcScore(score1, score2, score3, score4, score5);

然而,这会带来一个新问题:在编译时,编译器不知道名为calcScore的函数,该函数采用类型为double的5个参数。

您正在声明两个名为calcScore的函数:一个是简单声明的,没有定义没有参数,另一个是使用参数声明和定义的:

// just a declaration. This function has no definition and will not link
void calcScore();

// declared and defined
void calcScore( double score1, double score2, double score3, double score4,double score5)
{
}

您可以删除calcScore的第一个声明,因为我假设您不需要两者,并将其更改为:

void calcScore(double score1, double score2, double score3, double score4, double score5);

现在,当您尝试在calcScore中调用double并传入5 main时,编译器将知道您所指的功能。