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How to access a field of a struct by indexing?

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I have a 1-by-1 struct that possesses 3 fields named B, C, and D. Is there any way to call D by its index (i.e., D is the third field of struct A, so call the third field of struct A without mentioning the field name D) rather than its name (i.e, A.D)?
A.B = 1;
A.C = 2;
A.D = 3;
  1 Comment
Stephen23
Stephen23 on 26 Feb 2017
Edited: Stephen23 on 26 Feb 2017
@Rightia Rollmann: you might like to consider using a non-scalar structure, which lets you use indexing to access structures:

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Accepted Answer

Jan
Jan on 26 Feb 2017
Edited: Jan on 26 Feb 2017
A_cell = struct2cell(A);
D = A_cell{3}
Keep in mind that the order of the fields of structs is not necessarily constant:
A.B = 1;
A.C = 2;
A.D = 3;
B.B = 1;
B.D = 3;
B.C = 2;
isequal(A, B) % >> TRUE!
  4 Comments
Richard Crozier
Richard Crozier on 1 Aug 2019
If struct2cell results in copying the data in the structure, Guillaume's answer below is superior.
James Tursa
James Tursa on 1 Aug 2019
struct2cell creates shared-data-copies of the field variables. So, while there is overhead involved in creating the variable header info, the data itself is not copied.

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More Answers (1)

Guillaume
Guillaume on 26 Feb 2017
Yes, there is a way to get the nth field directly:
fns = fieldnames(A);
A.(fns{3})
But be aware that the order of the fields depends solely on the order in which they were created. As Jan pointed out, two structures may be indentical, yet have different field order.
Usually, you would only access fields by their index when you're doing some structure metaprogramming
  5 Comments
Vaishnavi
Vaishnavi on 19 Dec 2023
To add on to this question, can someone explain why A.(subsref(fieldnames(A),substruct('{}',{:}))) would not work?
Voss
Voss on 19 Dec 2023
@Vaishnavi: One reason that doesn't work is that you need single quotes around the colon in substruct. But even then, it may not do what you expect. Here's what it does (gets the value of the first field of A):
A = struct('field1',1,'field2',2)
A = struct with fields:
field1: 1 field2: 2
A.(subsref(fieldnames(A),substruct('{}',{':'})))
ans = 1
In order to know whether that "works", one would need to know what you expected it to do.

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