I’ve always thought that indexing with an empty array on the LHS of an = is a no-op for the variable on the LHS (there could be side effects of the evaluation of the RHS).
For example
% Case 1
A = [1 2;3 4];
A([]) = 5
But, assigning a non-scalar on the RHS result in an error
% Case 2
try
A([]) = [5,6];
catch ME
ME.message
end
Why doesn’t Case 1 result in an error insofar as it also had a different number of elements on the left and right sides?
Using the deletion operator on the RHS with an empty index on the LHS reshapes the matrix
% Case 3
A = [1 2;3 4];
A([]) = []
Why does a request to delete nothing from A change A?
But that doesn’t happen if A is column vector
% Case 4
A = [1;2;3;4];
A([]) = []
Deletion with a non-zero dimension of an empty index also reshapes the LHS
% Case 5
A = [1 2;3 4];
A(double.empty(0,1)) = []
A = [1 2; 3 4];
A(double.empty(1,0)) = []
I suppose Cases 3 and 5 are related to deleting a specific (non-empty) element from a matrix
% Case 6
A = [1 2;3 4];
A(3) = []
But I’m surprised that the result is not a column vector.
Are any/all of these results following a general rule? Pointers to relevant documentation would be welcome. I couldn’t find any.I’ve always thought that indexing with an empty array on the LHS of an = is a no-op for the variable on the LHS (there could be side effects of the evaluation of the RHS).
For example
% Case 1
A = [1 2;3 4];
A([]) = 5
But, assigning a non-scalar on the RHS result in an error
% Case 2
try
A([]) = [5,6];
catch ME
ME.message
end
Why doesn’t Case 1 result in an error insofar as it also had a different number of elements on the left and right sides?
Using the deletion operator on the RHS with an empty index on the LHS reshapes the matrix
% Case 3
A = [1 2;3 4];
A([]) = []
Why does a request to delete nothing from A change A?
But that doesn’t happen if A is column vector
% Case 4
A = [1;2;3;4];
A([]) = []
Deletion with a non-zero dimension of an empty index also reshapes the LHS
% Case 5
A = [1 2;3 4];
A(double.empty(0,1)) = []
A = [1 2; 3 4];
A(double.empty(1,0)) = []
I suppose Cases 3 and 5 are related to deleting a specific (non-empty) element from a matrix
% Case 6
A = [1 2;3 4];
A(3) = []
But I’m surprised that the result is not a column vector.
Are any/all of these results following a general rule? Pointers to relevant documentation would be welcome. I couldn’t find any. I’ve always thought that indexing with an empty array on the LHS of an = is a no-op for the variable on the LHS (there could be side effects of the evaluation of the RHS).
For example
% Case 1
A = [1 2;3 4];
A([]) = 5
But, assigning a non-scalar on the RHS result in an error
% Case 2
try
A([]) = [5,6];
catch ME
ME.message
end
Why doesn’t Case 1 result in an error insofar as it also had a different number of elements on the left and right sides?
Using the deletion operator on the RHS with an empty index on the LHS reshapes the matrix
% Case 3
A = [1 2;3 4];
A([]) = []
Why does a request to delete nothing from A change A?
But that doesn’t happen if A is column vector
% Case 4
A = [1;2;3;4];
A([]) = []
Deletion with a non-zero dimension of an empty index also reshapes the LHS
% Case 5
A = [1 2;3 4];
A(double.empty(0,1)) = []
A = [1 2; 3 4];
A(double.empty(1,0)) = []
I suppose Cases 3 and 5 are related to deleting a specific (non-empty) element from a matrix
% Case 6
A = [1 2;3 4];
A(3) = []
But I’m surprised that the result is not a column vector.
Are any/all of these results following a general rule? Pointers to relevant documentation would be welcome. I couldn’t find any. empty index, deletion, assignment MATLAB Answers — New Questions

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