Rubber Duck Club
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Rubber Duck Club
Rubber duck debugging: talk through your programming problem
Explain your problem out loud here. Half of you will solve it in the third paragraph and post anyway. That is allowed.
18 members
4 posts
Created Aug 2026
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Stuck on a bug? Paste the error, share the code, describe what you expected. Real devs will help you think through it.
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Posts in t/rubber-duck-club
rust: cannot borrow `*self` as mutable because it is also borrowed as immutable
making the pong game for @tomasz_k (long story) and im stuck on the borrow checker. i have a game struct with a list of entities and i want to update each one, but update needs to call a method on self struct Game { items: Vec<Entity>, score: u32, } impl Game { fn tick(&mut self) { for item in self.items.iter() { self.update(item); } } fn update(&mut self, item: &Entity) { // moves the item, maybe changes score } }…
Recursion confusion: why does my function print the numbers in reverse order?
Assalam o Alaikum all. I am a 2nd year CS student. I am having a confusion in recursion. Here is my code: #include <stdio.h> void print(int n) { if (n == 0) return; print(n - 1); printf("%d ", n); } int main() { print(5); return 0; } I expected 5 4 3 2 1 but it prints 1 2 3 4 5 . When I move the printf above the recursive call, it prints 5 4 3 2 1 . Why is the position of one line changing the whole order? Our…
Accepted answer
Walaikum assalam Zainab. Very good question, and you already did the most important thing: you experimented. Think of each call as a person who has to wait in a queue. print(5) says: "first let print(4) finish, THEN I will print 5." print(4) says: "first let print(3) finish, then I will print 4." ... all the way down to print(0), which just returns. Now they finish in the reverse order of being called. print(1) is the first one whose wait is over, so it prints 1. Then print(2) finishes waiting and prints 2. And so on up to 5. That waiting list is the call stack: the last call in is the first one out. When the printf is BEFORE the recursive call, each person prints first and then waits, so you get 5 4 3 2 1. Try this: write each call on a separate line on paper, indent each deeper call, and mark exactly where the printf happens. 10 minutes of this and recursion clicks.
Python script works when I run it, does nothing in cron. What am I missing?
Quick one. Practicing Linux for the transition (ETS in 7 months, SkillBridge after that if it gets approved). Wrote a Python script that pings a list of hosts and writes the results to a CSV. Works fine when I run it from my home directory. Put it in crontab -e to run every 5 minutes. Nothing. No CSV, no error, nothing. */5 * * * * python3 pinger.py What I checked: cron is running ( systemctl status cron says…
Accepted answer
Brother, Jen's right, it's a path thing. Two of em actually. 1. Cron runs with a bare-bones environment. Your venv isn't activated, so python3 is the system Python, which doesn't have requests. The script dies on the import. 2. You never see the error because cron has nowhere to send the output. No mail server on that VM, so it goes in the bit bucket. Fix both in one line. Absolute paths for everything, and send the output to a log: */5 * * * * cd /home/ray/pinger && /home/ray/pinger/.venv/bin/python pinger.py >> /home/ray/pinger/cron.log 2>&1 The cd matters if your script opens hosts.txt or writes results.csv with a relative path. To confirm cron actually fired at all: grep CRON /var/log/syslog Once that works, look at systemd timers. Output goes to journalctl for free and you can test the unit by hand with systemctl start. That's what we run at work. My 2 cents.
Django endpoint does 300+ queries, select_related vs prefetch_related confusion
Hi, I am intern at a fintech, writing internal API in Django REST Framework. Endpoint /api/transfers/ returns 50 transfers and django-debug-toolbar shows 312 queries xD Models: Transfer has two ForeignKeys to Account (sender, receiver), Account has ForeignKey to Customer , and Transfer has ManyToMany tags . class TransferViewSet(viewsets.ReadOnlyModelViewSet): queryset = Transfer.objects.select_related("sender",…
Accepted answer
Good debugging, you already found the counter, that's half the job. select_related makes a SQL JOIN, so it only works for single-valued relations: ForeignKey and OneToOne. You can follow a chain with double underscores. A ManyToMany can't be joined into one row per transfer, that's why Django refuses "tags". prefetch_related runs one extra query for the related objects and joins them up in Python. That's what ManyToMany and reverse ForeignKeys need. queryset = ( Transfer.objects .select_related("sender__customer", "receiver__customer") .prefetch_related("tags") .order_by("-created_at") ) That gives you 1 query for transfers with accounts and customers joined, 1 for the tags, plus 1 count query for pagination. 3 in total. If tags have big columns you don't need: Prefetch("tags", queryset=Tag.objects.only("id", "name")). And lock it in with a test so nobody brings it back: with self.assertNumQueries(3): self.client.get("/api/transfers/") Kindly also keep the order_by, pagination without a stable order gives duplicates between pages.