c++ - 无法将参数 ‘char*’ 的 ‘int*’ 转换为 ‘1’ 到 ‘int median(int*, int)’ array2 = median(array,size);

Question

我的问题正如我在标题中指定的那样:test_median.cpp:

In function ‘int main()’: test_median.cpp:26:27: error: cannot convert ‘char*’ to ‘int*’ for argument ‘1’ to ‘int median(int*, int)’

array2 = median(array,size);

这是我的代码。这是测试代码；

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

using namespace std;

int main()
{

int size;
cout<< "size gir" <<endl;
cin >> size;

int i;
char array[size];

cout<< "değerleri gir"<<endl;

for (i=0;i<size;i=i+1)
{
cin >> array[i];
}

char array2[size];

array2 = median(array,size);

cout<<"array2 = "<<array2<<endl;

return 0;
}

这是中值函数代码:

#include <iostream>
#include <cmath>

using namespace std;

int median(char array[], int size)
{
int i = 0;
int m,n;
while (i<size)
{
    if (array[i]>array[i+1])
    {
    m = array[i]; n = array[i+1];
    array[i] = n; array[i+1] = m;
    i=i+1;
    }
    else
    {
    i=i+1;
    }
}

return *array;

}

最后是“IpFunctions.h” header :

// this is just a headerfile that you add your Image processing functions' prototypes.
// when testing your functions with a main function
// main needs a prototype of the function

//Image medianfiltering(Image &inimg, const int size );
int median(int array[],int size);

// when you type a function add the header here !!! 
// such as 
// Image negative(Image &inimg);

所以我只想编写一个函数来获取数组的中位数并返回该数组。

Answer 1

在标题中

int median(int array[],int size);

在.cpp 文件中

int median(char array[], int size)

您的 main 正在尝试调用第一个(这是它唯一知道的)