c - 在 C 中寻找中位数

Question

我写这段代码是为了求均值和中位数，但我总是犯这些错误:

"71 warning: passing argument 1 of 'median' makes pointer from integer without a cast"

和:

"14 note: expected 'int *' but argument is of type 'int'".

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>

float mean(float x,int y)
{
float toplam = x;
int adet=y;

return toplam/adet;

}

int median(int AlinanSayilar[200],int adet)
{
int kacinci,kacinci2;
int medyan=0;

if(adet%2==1)
{
    kacinci=(adet/2)-1;
    kacinci2=kacinci+1;
    medyan=(AlinanSayilar[kacinci]+AlinanSayilar[kacinci2])/2;
}
else
{
    kacinci=(adet/2)-0.5;
    medyan=AlinanSayilar[kacinci];
}

printf("%d",medyan);

return 0;
}

int main()
{
int sayilar[200];
int i,k,j,holder;
float sum=0;

printf("Welcome the calculator...\n\tThis calculator finds mean,median and 
mode of your numbers...\n");
printf("\t\tNOTE:Please enter only integer numbers...\n\n");

for(i=0;true;i++)
{
    printf("Please enter a number(press -1 for exit): ");
    scanf("%d",&sayilar[i]);
    printf("\n");

    if(sayilar[i]==-1){
        break;
        }
    sum +=sayilar[i];
}

for(k=0;k<i-1;k++)
{
    for(j=k+1;j<i;j++)
    {
        if(sayilar[k]>sayilar[j])
        {
            holder=sayilar[k];
            sayilar[k]=sayilar[j];
            sayilar[j]=holder;
        }
    }
}

printf("Mean:%.2f",mean(sum,i));
median(sayilar[i],i);

system("pause");
return 0;
}

我该怎么办？在此先感谢您的帮助。如果您知道查找模式(最重复的数字)，您可以编写它的代码吗？

Answer 1

您的 median 函数接受一个数组和一个整数，但在您的 main() 函数中您使用两个整数调用它。 sayilar[i] 是 sayilar 数组的第 i 个索引，所以它是一个 int。

您可以通过将行更改为来修复它

median(sayilar,i);