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def test_args_2(x, y, z): print(x, y, z)

When calling that function with :

test_args_2(4, y=4)

Python gives:

TypeError: test_args_2() missing 1 required positional argument: 'z'

z is missing but it's considered positional not a keyword one, why ?

if we modify the call by this:

test_args_2(4, y=5, 4)

Python gives:

SyntaxError: positional argument follows keyword argument

Why z is considered positional although in the second error message it's assumed keyword one?

Salim Boukerma
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2 Answers2

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z will always be considered positional as long as you don't assign it a default value, for example:

def test_args_2(x, y, z=42):
    print(x, y, z)

You can read a more in-depth explanation here.

Atalajaka
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  • But is this acceptable ?, after a keyword argument no positional one can be used!, I mean z is supposed to become keyword because it is placed after y – Salim Boukerma Apr 08 '22 at 10:20
  • Yes, that is how python's function arguments work, as seen in the documentation: https://docs.python.org/3/tutorial/controlflow.html#more-on-defining-functions – Atalajaka Apr 08 '22 at 12:33
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def test_args_2(x, y, z):
    print(x, y, z)
test_args_2(4, y=4)  #not allow
test_args_2(x=4,y=2,z=5) #allow
test_args_2(y=2,z=5,x=4) #allow

Positional arguments are arguments that can be called by their position in the function call.
x,y,z are the positional arguments , all three are required

def test_args_2(x, y, z=0):
    print(x, y, z)
test_args_2(1,2) # allow

The error positional argument follows keyword argument means that the if any keyword argument is used in the function call then it should always be followed by keyword arguments. Positional arguments can be written in the beginning before any keyword argument is passed.

def test_args_2(x, y, z):
    print(x, y, z)
test_args_2(4, 5, 4) #allow
tst_args_2(2,3,z=4) # allow
test_args_2(4,y=5,z=3) #allow
test_args_2(x=2,y=3,z=5) #allow
tst_args_2(x=2,y=3,4) # not allow
tst_args_2(x=2,3,4) # not allow
tst_args_2(2,y=3,4) # not allow
zeeshan12396
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