为什么 function 返回 0？

Question

I have this function that should take a matrix, compare the diagonal elements of the matrix and find the smallest one.

Here it should compare y[0][1] and y[1][0] .

#include <iostream>
int Min(int, int [][2]);
using namespace std;
int main() {
   int min, y[2][2];
   y[0][0]=5;
   y[0][1]=4;
   y[1][0]=-9;
   y[1][1]=0;
   min = Min(2, y);
   cout<<min;
   return 0;
}  
int Min(int k, int x[][2]){
  int min=x[0][0];
  for(int i=0; i<k;i++){
   if(x[i][i]<min){
     min=x[i][i];
   }
  }
  return min;
}

It always returns 0. Why?

Answer 1

Here it should compare y[0][1] and y[1][0].

Your function goes through the diagonal of the matrix, it hence checks y[0][0] and y[1][1] . They are 5 and 0. The result is zero, which is to be expected.

Answer 2

Here it should compare y[0][1] and y[1][0] .

But that's not what you say here:

int min=x[0][0];

or here:

if(x[i][i]<min){
    min=x[i][i];
}

Since i cannot be both 0 and 1 at the same time, this accesses x[0][0] and x[1][1] .

And for those elements, 0 is the correct minimum.

As a side note, your Min function can only work with matrices of size 2, so the k parameter is unnecessary. Multi-dimensional arrays are annoying like that.

Answer 3

To iterate over all items instead of just [0][0] and [1][1] you need to do:

for(int i=0; i<k; i++)
{
   for(int j=0; j<k; j++)
   {
     if(x[i][j]<min)
     {
       min=x[i][j];
     }
   }
}

Answer 4

   int Min(int k,const int x[][2])
    {
      int min=x[0][0];
      for(int i=0; i<k;i++)
      {
        for(int j=0;j<k;j++)
        {
          if(x[i][j]<min)
           {
              min=x[i][j];
           }
        }
       }

Earlier it was always showing zero because it had a mess with the column index. Outer for loop iterates k-1 times with first iteration at i=0, and second at i=1 and during both iterations it assigns the same index to both row and column(ie, x [0][0] and x[1][1]). Perhaps it must assign the index x[0][0], x[0][1], x[1][0], x[1][1]. Earlier, it had only two iterations(the outer for loop) but now it takes four iterations assigning the appropriate index to both column and rows the efficient number of times.