为何Python无法检测arr.sort未加括号的语法错误，而Java可以？

Great question! This comes down to core differences between Python's dynamic nature and Java's static typing, along with how each language treats method references vs. actual method calls. Let's break it down step by step:

1. arr.sort is valid Python syntax (not an error) #

In Python, methods are first-class objects—meaning you can reference them, assign them to variables, or pass them around just like any other value. When you write arr.sort, you're not making a syntax mistake; you're simply grabbing a reference to the list's sort method itself, rather than executing it.

If you just run the line arr.sort on its own, Python will execute the statement (retrieve the method object) but won't perform any sorting. Since this is a completely legal operation, the interpreter won't throw an error—even though it doesn't do what you intended.

2. Java's static typing catches this at compile time #

Java is a statically typed language, which means the compiler checks for syntax and type correctness before the program ever runs. In Java, a method like sort() must be called with parentheses to execute it; writing arr.sort would be flagged immediately as a compile error, because the compiler knows sort is a method that requires invocation, not a variable you can reference directly.

Python has no compile-time checking for this kind of issue—it's a dynamic language, so it resolves types and operations at runtime. Since referencing a method is a valid runtime operation, no error is raised.

3. Why your program seemed to work #

Your input array was already in ascending order, so even without calling arr.sort(), the output looked correct. If you had tested with a shuffled array, you'd quickly notice that the sorting never happened, which would clue you in to the missing parentheses.

As a side note: if you had tried to use the result of arr.sort (e.g., sorted_arr = arr.sort), you'd end up with a method object stored in sorted_arr instead of a sorted list. Trying to use that variable later (like iterating over it) would throw an error—but on its own, arr.sort is harmless (just useless for your intended purpose).

内容的提问来源于stack exchange，提问作者Chris Ioannou