4 ms·

The problem isn't that simple use case. Although even in that case, they already had '=' as an assignment operator, and they could've easily kept it like the ma

The problem isn't that simple use case. Although even in that case, they already had '=' as an assignment operator, and they could've easily kept it like the majority of other languages do instead of introducing an inconsistency. The more major problem with the walrus operator is more complicated expressions they made legal with it. Like, could you explain to me why making these legal was a good thing? def foo() return ... def bar(): yield ... while foo() or (w := bar()) < 10: # w is in-scope here, but possibly nonexistent! # Even in C++ it would at least *exist*! print(w) # The variable is still in-scope here, and still *nonexistent* # Ditto as above, but even worse outside the loop print(w := w + 1) If they just wanted your use case, they could've made only expressions of the form 'if var := val' legal, and maybe the same with 'while', not full-blown assignments in arbitrary expressions, which they had (very wisely) prohibited for decades for the sake of readability. And they would've scoped the variable to the 'if', not made it accessible after the conditional. But nope, they went ahead and just did what '=' does in any language, and to add insult to injury, they didn't even keep the existing syntax when it has exactly the same meaning. And it's not like they even added += and -= and all those along with it (or +:= and -:= because apparently that's their taste) to make it more useful in that direction, if they really felt in-expression assignments were useful, so it's not like you get those benefits either.

DangitBobby 5y ago
In your example, if you leave out the parentheses around w := bar(), you get "SyntaxError: cannot use assignment expressions with operator" which makes me think it's a bug in the interpreter and not intentionally designed to allow it. I am baffled to learn that it's kept in scope outside of the statement it's assigned, and I agree it would have a negative impact on readability if used outside of the if statement.
- dataflow 5y ago
  > if you leave out the parentheses around w := bar(), you get "SyntaxError: cannot use assignment expressions with operator" which makes me think it's a bug in the interpreter and not intentionally designed to allow it. No, I'm pretty sure that's intentional. You want the left-hand side of an assignment to be crystal clear, which "foo() or w := bar()" is not. It looks like it's assigning to (foo() or w).
  - DangitBobby 5y ago
    To be clear: def thing(): return True if thing() or w:= "ok": # SyntaxError: cannot use assignment expressions with operator pass print(w) . . . if thing() or (w := "ok"): pass print(w) # NameError: name 'w' is not defined The first error makes me think your concern (that w is conditionally undefined) was anticipated and supposed to be guarded against with the SyntaxError. I believe the fact you can bypass it with parentheses is a bug and not an intentional design decision.
    - dataflow 5y ago
      Oh I see, you're looking at it from that angle. But no, it's intentional. Check out PEP 572 [1]: > The motivation for this special case is twofold. First, it allows us to conveniently capture a "witness" for an any() expression, or a counterexample for all(), for example: if any((comment := line).startswith('#') for line in lines): print("First comment:", comment) else: print("There are no comments") I have a hard time believing even the authors (let alone you) could tell me with a straight face that that's easy to read. If they really believe that, I... have questions about their experiences. The beauty of Python... [1] https://www.python.org/dev/peps/pep-0572/ https://www.python.org/dev/peps/pep-0572/
      - DangitBobby 5y ago
        Your new example makes me wonder: if I can intentionally conditionally bring variables into existence with the walrus operator, what's the motivation behind the SyntaxError in my statement above? I maintain my belief that the real issue here is, readability aside, if blocks do not implement a new scope, which has always been a problem in the language. The walrus operator just gives you new ways to trip over that problem. From the PEP: > An assignment expression does not introduce a new scope. In most cases the scope in which the target will be bound is self-explanatory: it is the current scope. If this scope contains a nonlocal or global declaration for the target, the assignment expression honors that. A lambda (being an explicit, if anonymous, function definition) counts as a scope for this purpose. I find this particularly strange and inconsistent: lines = ["1"] [(comment := line).startswith('#') for line in lines] print(comment) # 1 [x for x in range(3)] print(x) # NameError: name 'x' is not defined
- deleted 5y ago
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eesmith 5y ago
While the walrus operator gives a way to see this sort of non-C++ behavior, it's more showing that Python isn't C++ than something special about the operator. Here's another way to trigger the same NameError, via "global": import random def foo(): return random.randrange(2) def bar(): global w w = return random.randrange(20) return w while foo() or (bar() < 10): print(w) For even more Python-is-not-C++-fun: import re def parse_str(s): def m(pattern): # I <3 Perl! nonlocal _ _ = re.match(pattern, s) return _ is not None if m("Name: (.*)$"): return ("name", _[1]) if m("State: (..) City: (.*)$"): return ("city", (_[2], _[1])) if m(r"ZIP: (\d{5})(-(\d{4}))?$"): return ("zip", _[1] + (_[2] if _[2] else "")) return ("Unknown", s) del _ # Remove this line and the function isn't valid Python(!) for line in ( "Name: Ernest Hemingway", "State: FL City: Key West", "ZIP: 33040", ): print(parse_str(line))
- dataflow 5y ago
  Right, I'm quite well-aware of that, but I'm saying this change has made the situation even worse. If they ensured the variables were scoped and actually initialized it'd have actually been an improvement.
  - eesmith 5y ago
    I apologize. I misinterpreted your comment: # w is in-scope here, but possibly nonexistent! # Even in C++ it would at least *exist*! because I don't see how bringing up C++'s semantics is relevant when Python has long raised an UnboundLocalError for similar circumstances. If I understand you correctly, you believe Python should have introduced scoping so the "w" would be valid only in the if, elif, and else clauses, and not after the 'if' ends. This would be similar to how the error object works in the 'except' clause: >>> try: ... 1/0 ... except Exception as err: ... err = "Hello" ... >>> err Traceback (most recent call last): File "<stdin>", line 1, in <module> NameError: name 'err' is not defined If so, I do not have the experience or insight to say anything meaningful.

DangitBobby 5y ago

In your example, if you leave out the parentheses around w := bar(), you get "SyntaxError: cannot use assignment expressions with operator" which makes me think it's a bug in the interpreter and not intentionally designed to allow it. I am baffled to learn that it's kept in scope outside of the statement it's assigned, and I agree it would have a negative impact on readability if used outside of the if statement.

dataflow 5y ago
> if you leave out the parentheses around w := bar(), you get "SyntaxError: cannot use assignment expressions with operator" which makes me think it's a bug in the interpreter and not intentionally designed to allow it. No, I'm pretty sure that's intentional. You want the left-hand side of an assignment to be crystal clear, which "foo() or w := bar()" is not. It looks like it's assigning to (foo() or w).
- DangitBobby 5y ago
  To be clear: def thing(): return True if thing() or w:= "ok": # SyntaxError: cannot use assignment expressions with operator pass print(w) . . . if thing() or (w := "ok"): pass print(w) # NameError: name 'w' is not defined The first error makes me think your concern (that w is conditionally undefined) was anticipated and supposed to be guarded against with the SyntaxError. I believe the fact you can bypass it with parentheses is a bug and not an intentional design decision.
  - dataflow 5y ago
    Oh I see, you're looking at it from that angle. But no, it's intentional. Check out PEP 572 [1]: > The motivation for this special case is twofold. First, it allows us to conveniently capture a "witness" for an any() expression, or a counterexample for all(), for example: if any((comment := line).startswith('#') for line in lines): print("First comment:", comment) else: print("There are no comments") I have a hard time believing even the authors (let alone you) could tell me with a straight face that that's easy to read. If they really believe that, I... have questions about their experiences. The beauty of Python... [1] https://www.python.org/dev/peps/pep-0572/ https://www.python.org/dev/peps/pep-0572/
    - DangitBobby 5y ago
      Your new example makes me wonder: if I can intentionally conditionally bring variables into existence with the walrus operator, what's the motivation behind the SyntaxError in my statement above? I maintain my belief that the real issue here is, readability aside, if blocks do not implement a new scope, which has always been a problem in the language. The walrus operator just gives you new ways to trip over that problem. From the PEP: > An assignment expression does not introduce a new scope. In most cases the scope in which the target will be bound is self-explanatory: it is the current scope. If this scope contains a nonlocal or global declaration for the target, the assignment expression honors that. A lambda (being an explicit, if anonymous, function definition) counts as a scope for this purpose. I find this particularly strange and inconsistent: lines = ["1"] [(comment := line).startswith('#') for line in lines] print(comment) # 1 [x for x in range(3)] print(x) # NameError: name 'x' is not defined
deleted 5y ago
[deleted]

eesmith 5y ago

While the walrus operator gives a way to see this sort of non-C++ behavior, it's more showing that Python isn't C++ than something special about the operator. Here's another way to trigger the same NameError, via "global": import random def foo(): return random.randrange(2) def bar(): global w w = return random.randrange(20) return w while foo() or (bar() < 10): print(w) For even more Python-is-not-C++-fun: import re def parse_str(s): def m(pattern): # I <3 Perl! nonlocal _ _ = re.match(pattern, s) return _ is not None if m("Name: (.*)$"): return ("name", _[1]) if m("State: (..) City: (.*)$"): return ("city", (_[2], _[1])) if m(r"ZIP: (\d{5})(-(\d{4}))?$"): return ("zip", _[1] + (_[2] if _[2] else "")) return ("Unknown", s) del _ # Remove this line and the function isn't valid Python(!) for line in ( "Name: Ernest Hemingway", "State: FL City: Key West", "ZIP: 33040", ): print(parse_str(line))

dataflow 5y ago
Right, I'm quite well-aware of that, but I'm saying this change has made the situation even worse. If they ensured the variables were scoped and actually initialized it'd have actually been an improvement.
- eesmith 5y ago
  I apologize. I misinterpreted your comment: # w is in-scope here, but possibly nonexistent! # Even in C++ it would at least *exist*! because I don't see how bringing up C++'s semantics is relevant when Python has long raised an UnboundLocalError for similar circumstances. If I understand you correctly, you believe Python should have introduced scoping so the "w" would be valid only in the if, elif, and else clauses, and not after the 'if' ends. This would be similar to how the error object works in the 'except' clause: >>> try: ... 1/0 ... except Exception as err: ... err = "Hello" ... >>> err Traceback (most recent call last): File "<stdin>", line 1, in <module> NameError: name 'err' is not defined If so, I do not have the experience or insight to say anything meaningful.