0
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It's simple, simply write the shortest code to raise a TypeError.

In Python you can do:

raise TypeError

However it should be shorter with a code that gives a TypeError without raising it.

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7
  • 2
    \$\begingroup\$ this isn't really defined for other languages. You've also immediately answered the shortest possible way for Python... \$\endgroup\$
    – Jo King
    Jul 31, 2019 at 2:29
  • \$\begingroup\$ @JoKing I answer one out of 3 or more, feel free to post the other 3 byte ones \$\endgroup\$ Jul 31, 2019 at 2:30
  • 1
    \$\begingroup\$ This shouldn't be closed, but it's only well-defined for one language (Python) \$\endgroup\$
    – MilkyWay90
    Jul 31, 2019 at 2:31
  • 8
    \$\begingroup\$ I'm retracting my close vote, since this is pretty clear. But I'm downvoting this because it is an extremely simple challenge with an obvious optimal answer with no possibility of improvement. \$\endgroup\$
    – Jo King
    Jul 31, 2019 at 2:37
  • 3
    \$\begingroup\$ Oh right, you might want to post your challenges to the Sandbox before you post this here. \$\endgroup\$
    – user85052
    Jul 31, 2019 at 3:27

7 Answers 7

16
\$\begingroup\$
for chr1 in range(32, 128):
  for chr2 in range(32, 128):
    for chr3 in range(32, 128):
      code = chr(chr1) + chr(chr2) + chr(chr3)
      try:
        output = exec(code, {})
      except TypeError:
        print(code)
      except:
        pass

Try it online!

A naive search may result that you cannot trigger a TypeError within 2 characters. You may get a TypeError with 3 characters. All solutions including:

1. Math operator, non-numeric types

  • Math operator including +, -, ~
  • Non-numeric types including "", (), [], id, {}
+[]
-""
~id

2. Invoke number as function

0()

3. Matrix multiplication between numbers

@ is __matmul__ in python, read more here

0@0

4. Binary operator with float / complex number

~0.
~.0
~0j

5. Iterate over number

Thanks to @Dennis to point out this.

You may iterate some variable by star operator. And () for a tuple may be omitted.

*0,
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0
3
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JavaScript (V8), 4 bytes

1.()

Try it online!

[Output]

.code.tio:1: TypeError: 1 is not a function 1.() ^ TypeError: 1 is not a function at .code.tio:1:3

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2
\$\begingroup\$

Python 3, 3 bytes

+''

Try it online!

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2
  • 1
    \$\begingroup\$ Uh oh. I can not make my program less than 3 bytes. Anyway, nice trick! \$\endgroup\$
    – user85052
    Jul 31, 2019 at 2:31
  • \$\begingroup\$ @A__ Yeah, anyway you can post the other 3 byte ones \$\endgroup\$ Jul 31, 2019 at 2:31
2
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Python, 3 bytes

This is self-explanatory. The unary - operator does not take a string argument. This was covered in @tsh's solution above.

-""

Try it online!

Another one:

~id

Try it online!

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3
  • \$\begingroup\$ Haha yeah nice idea i still have more than 2 different styles in mind tho \$\endgroup\$ Jul 31, 2019 at 2:35
  • \$\begingroup\$ yeah, double quotes and that's about it \$\endgroup\$
    – Jo King
    Jul 31, 2019 at 2:36
  • \$\begingroup\$ There is also -id and +id \$\endgroup\$
    – Jo King
    Jul 31, 2019 at 2:38
2
\$\begingroup\$

Japt, 1 byte

Throws TypeError: U.í is not a function

í

Test it (Errors are displayed below the output field)


Or, a bit less trivial:

Japt, 2 bytes

Throws TypeError: (U++) is not a function

°(

Test it

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1
\$\begingroup\$

Python 3, 3 bytes

+[]

It gives an TypeError: bad operand type for unary +: 'list

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0
\$\begingroup\$

Python 3, 3 bytes

0()                                                              

raises a TypeError

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1
  • 1
    \$\begingroup\$ I just made a program that finds all the possible combinations for one and two bytes, execs that, and no two bytes raised a TypeError. Therefore, raising an TypeError with less than three bytes is impossible. \$\endgroup\$
    – forever
    Oct 27, 2020 at 17:22

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