Why does accessing a multi-dimensional array with fewer indices than its total dimensions not result in an error in MATLAB R2022b?

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I constructed a multi-dimensional array in MATLAB and called it with fewer indices. I was expecting to get an error message for the missing index, but the script was executed without any error message. Here's an example:
>> size(Pcc)
ans =
    5 6 5 6 6
>> Pcc(1,1,1,1)
ans =
    0.7854
Is this a bug?

Accepted Answer

MathWorks Support Team
MathWorks Support Team on 6 Nov 2023
Edited: MathWorks Support Team on 6 Nov 2023
If you access a multi-dimensional array in MATLAB with fewer indexes than its total dimensions, the array is treated as if it had only the dimensions you specified. The remaining dimensions are merged into a single dimension. In your case, the "Pcc" array has dimensions 5x6x5x6x6, and when you accessed it using only 4 indices, MATLAB collapsed the last two dimensions, 6x6, into a single dimension 36. As a result, when you execute the command "Pcc(1,1,1,1)", MATLAB accesses the "Pcc" array as if it were reshaped to "Pcc(5,6,5,36)" and returned the corresponding value at that specific location.
Here's an example to illustrate this behavior with a simpler 3-dimensional array:
>> A = rand(3, 4, 2)
>> A(2, 3)
ans =
    0.5469
>> B = reshape( A, 3, 8);
>> B(2, 3)
ans =
    0.5469
In this example, "A(2, 3)" has the same value with "B(2, 3)" where "B" is the same as array "A" reshaped to "A(3, 8)" where the last two dimensions are merged into a single dimension. 
Overall, this behavior is not a bug but rather a built-in feature of MATLAB. If you want to access specific elements in a multi-dimensional array, it is recommended to provide the appropriate number of indices for each dimension. 
  1 Comment
Bruno Luong
Bruno Luong on 2 Sep 2023
Edited: Bruno Luong on 2 Sep 2023
This answer not the question posted but rather a different one "Why does accessing a multi-dimensional array with more indices than its total dimensions not result in an error in MATLAB R2022b?"
The correct answer is @James Tursa answer bellow

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More Answers (1)

James Tursa
James Tursa on 1 Sep 2023
Edited: James Tursa on 1 Sep 2023
The accepted answer is misleading. When you access a multi-dimensional array with fewer indexes than its total dimensions, MATLAB treats the array as if it had only the dimensions you used, with the last dimension being all the actual remaining dimensions bunched into one dimension in the same memory order. E.g.,
A = reshape(1:24,2,3,4)
A =
A(:,:,1) = 1 3 5 2 4 6 A(:,:,2) = 7 9 11 8 10 12 A(:,:,3) = 13 15 17 14 16 18 A(:,:,4) = 19 21 23 20 22 24
A(2,3)
ans = 6
A(2,5)
ans = 10
So, why did that last A(2,5) work? Clearly 5 is greater than the actual 2nd dimension value 3. It worked because the (2,3,4) dimensions of A are treated as if the dimensions are (2,3*4) = (2,12) when accessing with only two indexes. The last part is simply accessed in memory order. It is as if you did a temporary reshape(A,2,12) first:
reshape(A,2,3*4)
ans = 2×12
1 3 5 7 9 11 13 15 17 19 21 23 2 4 6 8 10 12 14 16 18 20 22 24
And from this you can see that the A(2,5) element is indeed 10 when A is treated as a temporarily reshaped 2D matrix.
Bottom line is when you access a multi-dimensional array using fewer indexes than there are dimensions, MATLAB treats the array as a reshaped array using the number of dimensions that matches the number of indexes you used with the last dimension being the product of all trailing dimensions. Another example:
A = reshape(1:36,3,2,3,2)
A =
A(:,:,1,1) = 1 4 2 5 3 6 A(:,:,2,1) = 7 10 8 11 9 12 A(:,:,3,1) = 13 16 14 17 15 18 A(:,:,1,2) = 19 22 20 23 21 24 A(:,:,2,2) = 25 28 26 29 27 30 A(:,:,3,2) = 31 34 32 35 33 36
A(2,10)
ans = 29
reshape(A,3,2*3*2)
ans = 3×12
1 4 7 10 13 16 19 22 25 28 31 34 2 5 8 11 14 17 20 23 26 29 32 35 3 6 9 12 15 18 21 24 27 30 33 36
Again, you can see that the A(2,10) spot matches what you would expect from the reshaped matrix.
  2 Comments
Stephen23
Stephen23 on 2 Sep 2023
Edited: Stephen23 on 2 Sep 2023
Note that the indexing documentation does not mention this behavior, but it is explained here:
An interesting corollary of this is that linear indexing is just an edge-case of subscript indexing.
Haman
Haman on 6 Nov 2023
Thank you for pointing out the flaw in the posted answer. We've updated the answer to include your suggested change.

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