我正在写一个统计计算器，有 3 种不同的计算选项。问题是每当我选择第二个选项时，它都想打印第一个选项和第二个选项的答案。当我选择第 3 个选项时，它只会打印第 3 个选项的答案（错误的答案，但这可能是公式中的一个失误）。结果如下：
Please Enter a number of inputs
3
Please enter number 1
1
Please enter number 2
2
Please enter number 3
3
Statistical Calculator Menu
(1) Mean
(2) Standard Deviation
(3) Range
(4) Restart/Exit
2
Here is the Mean 2.0Standard Devition is 0.8
现在我认为这可能是我如何调用每个函数的问题，但我能说的最好的情况并非如此。然后我认为它可能是我没有初始化的值，但似乎也不是这样。我只需要另一双眼睛来看看我哪里出错了。
#include <stdio.h>
#include <conio.h>
#include <math.h>
const int MAX_DATA=8;
void menu(float numbers[], int amount);
float mean(float numbers[],int amount);
float standard_dev(float numbers[], int amount);
float range( float numbers[], int amount);
int main()
{
int i=0, amount=0;
float numbers[MAX_DATA];
printf("Please Enter a number of inputs \n");
scanf("%d", &amount);
if (amount>MAX_DATA)
{
printf("You entered too many numbers");
}
else
{
for (i=0; i<amount; i++)
{
printf("Please enter number %d\n", i+1);
scanf("%f",&numbers[i]);
}
menu(numbers,amount);
}
getch();
return 0;
}
void menu(float numbers[],int amount)
{
int input2=0;
printf("Statistical Calculator Menu");
printf("\n(1) Mean\n(2) Standard Deviation\n(3) Range\n(4) Restart/Exit\n");
scanf("%d",&input2);
if(input2==1)
{
mean(numbers,amount);
}
if (input2==2)
{
standard_dev(numbers,amount);
}
if (input2==3)
{
range(numbers,amount);
}
}
float mean(float numbers[],int amount)
{
int i;
float sum=0;
float average=0;
for (i=0; i<amount; i++)
{
sum=sum+numbers[i];
}
average=sum/amount;
printf("Here is the Mean %.1f", average);
return average;
}
float standard_dev(float numbers[], int amount)
{
float sdev=0,dev=0,sumsqr=0,variance=0;
int i;
float mean2=0;
mean2=mean(numbers,amount);
for (i=0; i<amount; i++)
{
dev=numbers[i]-mean2;
sumsqr+=dev*dev;
}
variance=sumsqr/(float)amount;
sdev=sqrt(variance);
printf("Standard Devition is %.1f", sdev);
return sdev;
}
float range(float numbers[],int amount)
{
int i;
float diff=0;
for (i=0; i<=amount; i++)
{
diff=numbers[amount]-numbers[1];
}
printf("%f\n",diff);
return diff;
}