Int(...) @Error Usage

[confucion]

Is this usage correct?

While 1
  $tmp = Int("abc")
  If @error = 1 Then ExitLoop
Wend

The loop is exited upon first run?

[jvanegmond]

Is this usage correct?

While 1
  $tmp = Int("abc")
  If @error = 1 Then ExitLoop
Wend

The loop is exited upon first run?

Usage is correct. Behavior is also correct. Int tries to turn "abc" into an integer. Since it can't make any sense of it (there are no numbers, could you make sense of it?) it will error. Upon error your script exits. Edited February 4, 2011 by Manadar

[Melba23]

confucion,

Why should it not exit?

The string "abc" is not an integer and so the function will return 0 and set @error to 1 just as the Help file says.

If you are referring to the remainder of the Help file line where it says "if not an integer, float or string" then you need to understand that the function will always fail unless the value is in integer, float or string format.

Look at the results from this:

ConsoleWrite(Int(1) & @CRLF)    ;integer format
ConsoleWrite(Int(1.0) & @CRLF)  ; float format
ConsoleWrite(Int("1") & @CRLF)  ; string format

You can see that they all return 1 saying that the item is an integer.

All clear?

M23

[confucion]

Because when I execute the code above, my CPU is stuck at 100% and the script doesn't quit indicating it is in infinite loop.

Have you tried running the above code?

[jvanegmond]

Because when I execute the code above, my CPU is stuck at 100% and the script doesn't quit indicating it is in infinite loop.

Have you tried running the above code?

You're right. It can convert any string format. If there are no numbers available then 0 is assumed. One of the things it fails at if you try to pass a pointer type:

While 1
  $tmp = Int(Ptr(0))
  If @error = 1 Then ExitLoop
Wend

[Melba23]

confucion,

That caught us all out!

Int failing on a string was reported as a bug (#1592) and dismissed as "No Bug". The trick is to do this:

$sString = "abc"
While 1
    If Not (IsNumber($sString) And Int($sString)) Then ExitLoop
Wend

Now you only test for an integer if the string is recognised as a number.

M23

[confucion]

Just found out I was using Int incorrectly to validate if the variable value was a string or number.

Use IsNumber instead in these situations.

While 1
  If IsNumber("abc") = 0 Then ExitLoop
Wend

This works.