C++ Array swapping and find smallest element - c++

Question

I am trying to find first smallest array but my code does not display any output. There are no errors or warnings. Actually, I am trying to check an algorithm that I got as an assignment from my university.
#include <iostream>
using namespace std;
int main(){
int arr[7]= {8,4,6,9,2,3,1};
int n = sizeof(arr)/sizeof(arr[0]);
int smallest = 0;
for(int j = 1; j = (n-1); j = (j + 1) )
{
smallest= j ;
for(int i = (j+1); i = n ; i = (i + 1))
{
if (arr[i]<arr[smallest])
{
smallest = i;
int swaper = arr[j];
arr[j] = arr[smallest];
arr[smallest] = swaper;
}
}
}
for(int a = 1; a = n; a = (a + 1))
{
cout<<arr[a];
}
return 0;
}

Answer 1

There are three errors with this code:
for(int a = 1; a = n; a = (a + 1))
{
cout<<arr[a];
}
Firstly, arrays start from zero, not one. So the first part of the for statement should be int a = 0;.
Secondly, you are not comparing a and n, you are assigning n to a, (and the value is non-zero, so you always keep going). The equality test is ==, but you don't want that anyway!
Thirdly, the loop condition is for when to keep going, not when to stop. So you need either < or != (either will work, people have long arguments about which is preferable).
The normal way to write a loop over a range of integers in C++ is:
for (int a = 0; a < n; a++)
You are at least consistent, and have made the same mistake in every loop. You will need to fix it in every loop.

Answer 2

# include <iostream>
using namespace std;
int main ()
{
int a[100][100],n,k,i,j,aux,mi=0;
cin>>n>>k;
for(i=1;i<=n;i++)
for(j=1;j<=k;j++)
cin>>a[i][j];
for(i=1;i<=n-1;i++)
for(j=i+1;j<=n;j++)
if(a[i][k]>a[j][k])
{aux=a[i][k];
a[i][k]=a[j][k];
a[j][k]=aux;
} //until here you are sorting the 2D array
for(i=1;i<=n;i++) {
for(j=1;j<=k;j++) {
cout<<a[i][j]<<" ";
}
cout<<endl;
}
cout<<endl;
mi=a[1][1];
for (i=1; i<=n; i++)
{
for (j=1; j<=n; j++)
if (mi<a[i][j])
mi=a[i][j];
} //here you're finding the smallest element
cout<<mi;
return 0;
}
The Code doesn't compile, but the idea should solve 90%, you just have to write the code.