This is a question about the behavior of fn!index() function. The syntax is fn!i

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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  • Hi Chetan,
    fn!index({4,5,6}, 1000, {})
    {}(default value)--> The default value to be returned if the array or the index is invalid, i.e. if array itself is empty, or index is < 1 or > array.length.
    "The type of the default value must be same as that of elements of the array."
    Thatsway fn!index({4,5,6}, {2,1000}, {})---4,5 display two values

    for better understand
    index({4,5,6}, 1000, -9)-result is -9

    I think it will help u
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  • Hi Chetan,
    fn!index({4,5,6}, 1000, {})
    {}(default value)--> The default value to be returned if the array or the index is invalid, i.e. if array itself is empty, or index is < 1 or > array.length.
    "The type of the default value must be same as that of elements of the array."
    Thatsway fn!index({4,5,6}, {2,1000}, {})---4,5 display two values

    for better understand
    index({4,5,6}, 1000, -9)-result is -9

    I think it will help u
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