CODEWITHSUNDEEP
functionidl-programming-language
SilvanD
SilvanD

Reputation: 325

How to accept arrays in my defined function

When I try to feed the following function an array, it gives me the following error: % Expression must be a scalar or 1 element array in this context: . How to modify such that I can either give it a scalar or array? 

; return integer -1, 0, or 1, depending on whether x is less than 0, equal to 0, or greater than 0, respectively
function whatisit, x
  case 1 of
    (x lt 0): y=-1
    (x eq 0): y=0
    (x gt 0): y=1
  endcase
  return, y
end

Upvotes: 0

Views: 855

Answers (1)

mgalloy
mgalloy

Reputation: 2386

This should do it:

function whatisit, x
  return, x gt 0 - x lt 0
end

EDIT: For pedagogical reasons, I will show the ugly (untested) looping solution, but you should never do this in IDL:

function whatisit, x
  n = n_elements(x)
  result = bytarr(n)
  for i = 0L, n - 1L do begin
    case 1 of
      x[i] lt 0: result[i] = -1
      x[i] eq 0: result[i] = 0
      x[i] gt 0: result[i] = 1
    endcase
  endfor
  return, size(x, /n_dimensions) eq 0 ? result[0] : result
end

Upvotes: 2

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