C++ 2d array row swap - c++

Question

Don't know if it matter but for the problem I use rand() to fill a 2d array with elements and CodeBloks for IDE.
I've been learning C++ and have gotten to pointers and 2d arrays and I've encountered something I don't understand why it works like this.
I had a tasks to swap two rows of a 2d array using pointers and this is how I swap the rows:
int *temp=arr[0];
arr[0]=arr[1];
arr[1]=temp;
And when my array is defined like this the code can compile without problem:
int a=5;
int b=3;
int **arr=new int*[a];
for(int i=0;i<a;i++){
*(arr+i)=new int[b];
for(int j=0;j<b;j++){
*(*(arr+i)+j)=rand()%(125-15+1)+15;
cout<<*(*(arr+i)+j)<<" ";
}
cout<<endl;
}
But when its defined like this the code doesn't compile doesn't and I don't understand what is the difrence:
int a=5;
int b=3;
int arr[a][b];
And I get this as an error message:
invalid array assignment|
incompatible types in assignment of 'int*' to 'int [b]'|
Aren't the two definitions equvelent trought the compiler or does something else happen?

Answer 1

Arrays are not pointers, and pointers are not arrays.
With
int **arr=new int*[a];
arr is not a two-dimensional array but a pointer to the first element in an array of pointers, and your code is only swapping pointers.
But you can't assign between arrays;
int a[3] {1,2,3};
int b[3] {4,5,6};
a = b; // Error
or from a pointer to an array;
int x = 0;
int a[3] {1,2,3};
a = &x; // Error
However, an array can be implicitly converted into a pointer to its first element, and when you use arrays your "swapping code" is equivalent to
int *temp = &arr[0][0]; // Pointer to the first element of the array arr[0]
arr[0] = &arr[1][0]; // Try to assign a pointer to the first element of arr[1] to the array arr[0]
arr[1] = temp; // Try to assign a pointer to the array arr[1]
In summary, you can't swap arrays, you need to do the swapping element-wise.
Or use std::array instead of "native" arrays, or std::vector if you need to determine the dimensions at runtime.