为什么调用函数参与运算再给变量赋值，数就不对了

Question

求线性回归方程a和b值，结果明显不对
代码如下

#include
//调用库函数
int main() //主函数
{ float aver(float a[],int); //声明求均值函数

float Temp_X[96] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1.46667, 11.4667, 31.6, 52.7333, 80.3333, 116.333, 156.6, 199.4, 242.2, 283.4, 329.2, 379.333, 431.333, 482.6, 541, 594.4, 643.533, 692.133, 736.267, 772.667, 810.133, 841.867, 868.2, 892.4, 917.667, 939.8, 954.667, 969, 976.8, 983.4, 987.467, 994.933, 1023.67, 875.2, 873.933, 758.8, 678.2, 515.867, 782.533, 908.8, 779.2, 831.4, 645.533, 734.067, 679.533, 610.267, 565.067, 512.467, 462, 405.2, 354.133, 302, 247.8, 191.533, 140, 94.2667, 57.5333, 25.9333, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 };            //定义初始化数组X 
float Temp_Y[96] = {0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 1.785, 2.57833, 3.927, 5.79233, 7.379, 9.48133, 11.1473, 12.4167, 13.6627, 16.193701, 18.248699, 19.042, 19.042, 19.105301, 16.6383, 17.240999, 14.631, 11.8217, 11.663, 12.155, 15.488, 21.859301, 19.32, 19.042, 19.6133, 21.105, 22.9937, 20.827299, 23.858299, 23.0333, 19.2883, 15.6937, 21.5893, 23.802999, 20.518299, 21.5893, 17.907301, 17.971001, 17.574301, 16.781, 15.5513, 12.3773, 10.2747, 8.60867, 6.86333, 5.39567, 3.88767, 2.856, 2.142, 2.142, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952 }; //定义初始化数组Y 
float a,b,sum1=0,sum2=0;                         //定义变量，a，b为线性回归方程y=ax+b中a，b的值，sum1，sum2，为求a时分步求解用的变量 
int i,j;                                              //i，j为循环时使用的变量 
for(i=0;i<96;i++)
{
    sum1+=(Temp_X[i]*Temp_Y[i]);
    sum2+=(Temp_X[i]*Temp_X[i]);
}
a=(sum1-96*aver(Temp_X,96)*aver(Temp_Y,96))/(sum2-96*aver(Temp_X,96)*aver(Temp_X,96));  //通过线性回归方程公式求出a值 
b=aver(Temp_Y,96)-(a*aver(Temp_X,96));
printf("a=%f,b=%f",a,b); 

}
float aver(float a[],int n) //求均值的函数
{ //用for循环求出所有值的和，再除以个数n
int i;
float sum;

for(i=0;i<n;i++)
{
sum+=a[i];
}
return sum/n; //返回结果
}

Answer 1

float aver(float a[],int n) { //用for循环求出所有值的和，再除以个数n 
    int i;
    float sum;
//  float sum = 0; // 要赋初值

    for(i=0;i<n;i++) 
        sum+=a[i];

    return sum/n; //返回结果 
}

这个函数sum没有赋初值

#include <stdio.h> 

#define N 96

int main() { 
    float Temp_X[N] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1.46667, 11.4667, 31.6, 52.7333, 80.3333, 116.333, 156.6, 199.4, 242.2, 283.4, 329.2, 379.333, 431.333, 482.6, 541, 594.4, 643.533, 692.133, 736.267, 772.667, 810.133, 841.867, 868.2, 892.4, 917.667, 939.8, 954.667, 969, 976.8, 983.4, 987.467, 994.933, 1023.67, 875.2, 873.933, 758.8, 678.2, 515.867, 782.533, 908.8, 779.2, 831.4, 645.533, 734.067, 679.533, 610.267, 565.067, 512.467, 462, 405.2, 354.133, 302, 247.8, 191.533, 140, 94.2667, 57.5333, 25.9333, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 };            //定义初始化数组X 
    float Temp_Y[N] = {0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 1.785, 2.57833, 3.927, 5.79233, 7.379, 9.48133, 11.1473, 12.4167, 13.6627, 16.193701, 18.248699, 19.042, 19.042, 19.105301, 16.6383, 17.240999, 14.631, 11.8217, 11.663, 12.155, 15.488, 21.859301, 19.32, 19.042, 19.6133, 21.105, 22.9937, 20.827299, 23.858299, 23.0333, 19.2883, 15.6937, 21.5893, 23.802999, 20.518299, 21.5893, 17.907301, 17.971001, 17.574301, 16.781, 15.5513, 12.3773, 10.2747, 8.60867, 6.86333, 5.39567, 3.88767, 2.856, 2.142, 2.142, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952 }; //定义初始化数组Y 

    float xSum = 0,ySum = 0, 
    float xySum=0, x2Sum=0;//这里赋值了后面就忘了 
    float a, b;

    for( int i = 0; i < N; i++ ) {
        xySum += (Temp_X[i] * Temp_Y[i]);
        x2Sum += (Temp_X[i] * Temp_X[i]);
        xSum += Temp_X[i];
        ySum += Temp_Y[i];
    }
//  a = (xySum - 96 * xAver * yAver) / (x2Sum-96 * xAver * xAver);  
    a = (N * xySum - xSum * ySum)/(N * x2Sum - xSum*xSum);
    b = ySum / N - a * xSum / N;
    printf( "a = %f\n b = %f", a, b ); 
}

最小二乘法