Why Does My Thread Execution Anomalous When I Don't Explicitly Invoke Threads in Python?

Thread Execution Anomalies: Demystifying Premature Thread Invocation

When creating threads in Python, it's crucial to invoke them explicitly using the start() method. However, in situations where the target function is invoked within the thread creation syntax, an intriguing issue arises.

The Problem: Running Threads Without Explicit Invocation

Consider the following code snippet:

t1 = threading.Thread(target=self.read())
print("something")
t2 = threading.Thread(target=self.runChecks(), args=(self,))

Surprisingly, the print statement never executes because self.read runs indefinitely, preventing the program from reaching the next line. This behavior seems counterintuitive, as calling t1.start() should initiate thread execution and allow the program to proceed.

The Solution: Understanding the Parenthesis Misconception

The issue lies in the trailing parentheses after self.read(). In Python, parentheses immediately following a function invoke it, so the following code:

target=self.read()

actually calls self.read and passes its return value as the target argument to Thread. However, Thread expects a function reference, not a return value. To correct the behavior, simply remove the parentheses and explicitly invoke t1.start() after thread creation:

t1 = threading.Thread(target=self.read)
t1.start()
print("something")

Handling Arguments in Thread Targets

When the target function requires arguments, use the args or kwargs arguments to threading.Thread. Alternatively, employ a lambda function, as shown below:

thread = threading.Thread(target=f, args=(a, b), kwargs={'x': c})

or

thread = threading.Thread(target=lambda: f(a, b, x=c))

Remember, if using a lambda, the function's arguments are evaluated when the lambda is invoked, not when it's defined. This can lead to unexpected results if variables are reassigned before thread scheduling.

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