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This is a question about the behavior of fn!index() function. The syntax is fn!i
chetany
A Score Level 1
over 8 years ago
This is a question about the behavior of fn!index() function. The syntax is fn!index(array, indices, defaultValToReturn).
The defaultValToReturn is of AnyType. Documentation mentions that default value should be of the same type as array elements. I find the behavior of index function strange in the below case:
fn!index( {4,5,6}, 1000, {}) -- As expected it returns a empty list i.e List of Variant since array size is less than 1000.
Now try, fn!index( {4,5,6}, {1000}, {} ) - This returns 4 - the first element in the array. I did not expect this result.
Now try, fn!index( {4,5,6}, {1000}, null ) - This returns nothing as expected.
Why does it behave in such manner for case 2? It always returns the first element whenever a multiple indices are provided and the default value is also an empty list.
Also try: fn!index({4,5,6}, {2,1000}, {}) - this returns 5,4. The 4 is the first element of the array.
I...
OriginalPostID-184786
OriginalPostID-184786
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chetany
A Score Level 1
over 8 years ago
@bhanuprakashm, the documentation does mention that the type of the default value must be the same as that of the elements of the array. However, it seems that internally the type of the default value and the type of the array elements is not compared when the index function executes. You can pass a default value which has a different data-type than the array elements. This maybe the cause of above strange behavior.
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chetany
A Score Level 1
over 8 years ago
@bhanuprakashm, the documentation does mention that the type of the default value must be the same as that of the elements of the array. However, it seems that internally the type of the default value and the type of the array elements is not compared when the index function executes. You can pass a default value which has a different data-type than the array elements. This maybe the cause of above strange behavior.
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