CODEWITHSUNDEEP
pythongeneratoryield
user3090226
user3090226

Reputation: 41

Why can't a yield expression be a function argument?

I simply want to tidy up the following code.

def well_known_generator():
    print('started')
    received = yield 100
    print(received)
    received = yield 200
    print(received)
    received = yield 300
    print('end')

g = well_known_generator()
print(next(g), g.send(None), g.send(None), g.send(None))

I just moved the yield expression into the print function, but a syntax error occurs. I'm just wondering why a yield expression can't be a function argument like below? If yield works like an expression, it should be okay as a function argument.

def well_known_generator():
    print('start')
    print(yield 100)
    print(yield 200)
    print(yield 300)
    print('end')

g = well_known_generator()
print(next(g), g.send(None), g.send(None), g.send(None))

SyntaxError: invalid syntax (<ipython-input-58-bdb3007bb80f>, line 3) 
  File "<ipython-input-58-bdb3007bb80f>", line 3
    print(yield 100)
              ^
SyntaxError: invalid syntax

Upvotes: 3

Views: 472

Answers (1)

Martijn Pieters
Martijn Pieters

Reputation: 1121266

You need to add another pair of parentheses around yield ...:

def well_known_generator():
    print('start')
    print((yield 100))
    print((yield 200))
    print((yield 300))
    print('end')

The parentheses are part of the yield expression syntax:

yield_atom       ::=  "(" yield_expression ")"

but the parentheses are optional when it's the only expression in an assignment statement or a statement expression:

The parentheses may be omitted when the yield expression is the sole expression on the right hand side of an assignment statement.

Inside a call expression (such as print(...)), the parentheses can't be omitted.

Upvotes: 8

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