What is the logic of python * syntax when iterating over a tuple sequence?

The

code is taken from page 4 of the third edition of pythoncookbook. The code is as follows: 

records =[
    ("foo", 1, 2),
    ("batr", "hello"),
    ("foo", 3 ,4),
]
def do_foo(x,y):
    print("foo", x ,y)

def do_bar(s):
    print("bar", s)

for tag, *args in records:
    if tag == "foo":
        do_foo(*args)
    elif tag == "bar":
        do_bar(*args)
:
foo 1 2
foo 3 4

the principle of the code is not very clear. Please explain it. Thank you.

Python
Mar.21,2021

I guess your confusion may be why the: bar hello results didn't show up?
because the second item of your records is misspelled: ('batr',' hello') should be ('bar',' hello'), 

In [1]: records =[
   ...:     ('foo', 1, 2),
   ...:     ('bar', 'hello'),
   ...:     ('foo', 3 ,4),
   ...: ]
   ...: def do_foo(x,y):
   ...:     print('foo', x ,y)
   ...: 
   ...: def do_bar(s):
   ...:     print('bar', s)
   ...: 
   ...: for tag, *args in records:
   ...:     if tag == 'foo':
   ...:         do_foo(*args)
   ...:     elif tag == 'bar':
   ...:         do_bar(*args)
   ...:         
foo 1 2
bar hello
foo 3 4

the simple explanation is:
* arg decomposes several elements after tag.
for example, the three elements in the list record are: ('foo', 1 hello' 2) (' bar', 'hello') (' foo', 3, 4)
tag is' element after foo', (1 br 2) is * args, so print foo 1 2
tag as' element behind bar', 'hello' is * args, So print bar hello
tag as' foo', followed by the element (3) is * args, so print foo 3 4

< H2 > use * to process the remaining elements < / H2 >

-- page 25 of "smooth python"

in Python, it is a classic way for a function to use * args to get an uncertain number of parameters.
in Python 3, this concept extends to parallel assignments: 

>>> a, *body, c, d = range(5)
>>> a, body, c, d
(0, [1, 2], 3, 4]
>>> *head, b, c, d = range(5)
([0, 1], 2, 3, 4]

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