Difference between the given syntax

Hi All,

I've doubt in interpreting following lines . Can somebody clarify that ?

cube.id(i)=lcount

cube(1).id=lcount1

Any help will be appreciated.

Thanks in advance!

Answer 1

[Ameer Hamza]

In case of

cube.id(i)=lcount

'cube' is a scalar struct and id is an array. And In case of

cube(1).id=lcount1

'cube' is a struct array, while 'id' is a scalar.

Answer 2

[Steven Lord]

As Ameer Hamza said, the difference is whether you want your struct array to be a scalar struct array with a field that is not scalar or whether you want a nonscalar struct array with fields that are scalar. Let's look at a slightly different example.

>> C1.id(2) = 42;
>> C2(2).id = 99;
>> whos C1 C2
  Name      Size            Bytes  Class     Attributes

  C1        1x1               184  struct              
  C2        1x2               184  struct              

C1 and C2 have different sizes.

>> disp(C1)
    id: [0 42]

C1 only has one element, that element has one field, and that field has two elements.

>> disp(C2)
1×2 struct array with fields:
    id

>> disp(C2(1))
    id: []
>> disp(C2(2))
    id: 99

C2 has two elements, each of those elements have one field, and the second of those elements has a field with one element.

One key question you should ask when deciding between a struct containing an array or an array of structs is how are you likely to want to "slice" your data. In C1, I can easily operate on each element of id with a loop, but if it had another field say name I would have to operate on each field separately. In C2 operating on all the id fields will be a bit more challenging, but I can easily extract both the id and the name for an individual element.

>> C1.name = ["Cleve", "Jack"];
>> C2(1).name = "Cleve";
>> C2(2).name = "Jack";
>> C2(1).id = 0;

% Display both ID and name from one of the people stored in C1
>> disp(C1.id(1))
     0
>> disp(C1.name(1))
Cleve

% Display both ID and name from one of the people stored in C2
>> disp(C2(1))
      id: []
    name: "Cleve"

% Display both ID values for people in C1
>> disp(C1.id)
     0    42
     
% Display both ID values for people in C2 -- note the square brackets
>> disp([C2.id])
     0    99