C ++标准偏差分配需要帮助纠正错误

时间:2015-11-11 22:35:26

标签: c++ standards standard-deviation

尝试编写使用数组计算标准偏差的程序。 它一直给我一个错误:

  

" LNK2019未解析的外部符号_main在函数" int中引用   __cdecl invoke_main(void)" (?invoke_main @@ YAHXZ)ConsoleApplication7777 C:\ Users \ Gregory \ Desktop \ ConsoleApplication7777 \ ConsoleApplication7777 \ MSVCRTD.lib(exe_main.obj)1"

这是我的代码:

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

using namespace std;

double standard_Deviation(double x[], int n);

//Main Function here
void Main()
{
    //declare variables here
    double x[100];
    double sDeviation;
    int i;
    int n;

//input number values here
cout << "Enter n value";
cin >> n;

//input array values here
cout << "Enter values:" << endl;

for (i = 0; i < n; i++)
    cin >> x[i];
//call standard deviation function
sDeviation = standard_Deviation(x, n);

//outputting standard deviation
cout << "Standard Deviation:" << sDeviation << endl;
//give it time to think
system("pause");

}

double standard_Deviation(double x[], int n)
{
double sd = 0;
double mean=0;
for (int i = 0; i<n; i++)
    mean = mean + x[i];
mean = mean / n;
for (int i = 0; i<n; i++)
    sd = sd + pow((x[i] - mean), 2);
sd = sd / n;
sd = sqrt(sd);
return sd;
return mean;
} //end of standard deviation

2 个答案:

答案 0 :(得分:1)

参加C和C ++的区分大小写;链接器寻找的主程序是

int main() {/**/}

void Main(){/* ... */}

答案 1 :(得分:0)

替换

void Main()

int main()

应用程序需要main函数(具有int返回类型)并且这是链接器抱怨的内容,因为您没有提供一个由于您的大写字母错误{{1} }。