Base CS from zero
Functions and the call stack: multiple-choice review
Crux Multiple-choice synthesis across the unit: calls, stack frames, return addresses, parameter passing, scope and lifetime, and what causes a stack overflow.
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You are at middle altitude — in the skySix questions that cut across the whole unit. Each one asks you to reason about what the machine actually does when a function is called — not to recite a definition, but to predict the state of the stack, the path of control, and where data lives.
Goal
Confirm you can connect the call mechanism, the stack frame, parameter passing, scope and lifetime, and stack overflow into one coherent model of how a function call works at the machine level.
Quiz
Completed
A function body lives at one address. The same function is called from three different places in the program. How does each call return to the right caller?
Heads-up A CALL is a jump to one fixed address, not a copy-paste. The body exists exactly once. What differs between call sites is the return address saved at call time, not the body.
Heads-up A fixed return target would only serve one caller. The whole point of saving a return address per call is that the same body can serve callers from anywhere and return to the correct one each time.
Heads-up There is one body at one address regardless of how many names or call sites refer to it. Returning correctly is handled at runtime by the saved return address, not by compile-time renaming.
Quiz
Completed
main calls outer, outer calls inner. While inner is executing, what is on the stack and in what order?
Heads-up A caller's frame is not popped when it calls another function; it waits below. main and outer are still active and still on the stack while inner runs. All three coexist.
Heads-up main was pushed first and sits at the bottom; inner was pushed last and sits on top. The newest frame is always on top, and it is popped first.
Heads-up Each call gets its own separate frame with its own return address and locals. They are distinct blocks stacked on top of one another, not merged.
Quiz
Completed
A function add(x, y) is called as add(3, 5). Where do the values 3 and 5 live while add runs, and what does that guarantee?
Heads-up A callee cannot name the caller's locals — scope forbids it. Arguments are copied into the callee's frame at the call, so add works on its own cells, not the caller's.
Heads-up Arguments are not globals. They are copied into the new frame, which is private to that call and discarded on return. Nothing global is involved.
Heads-up For primitive numbers the values themselves are copied (pass-by-value). Reassigning x inside add changes only add's cell, leaving the caller's argument untouched.
Quiz
Completed
Function f declares a local variable result. After f returns, code in main tries to use result. What is true?
Heads-up Locals do not migrate anywhere on return. The frame is popped and its cells are freed for reuse. There is no automatic hand-off of a callee's locals to its caller.
Heads-up The cells are reused by the next call almost immediately. Even before they are overwritten, the name is out of scope — reaching those cells would be a use-after-free bug, not valid access.
Heads-up After f returns, f's frame is gone, not below main's. And even while a caller's frame is on the stack, scope rules forbid naming another function's locals.
Quiz
Completed
countdown(n) returns when n < 0 and otherwise calls countdown(n - 1). What is the maximum number of frames on the stack at once for countdown(3)?
Heads-up A recursive call is a new CALL to the same address; it pushes a brand-new independent frame with its own n. Frames are not reused — they accumulate until the base case unwinds them.
Heads-up The base-case call countdown(-1) is still a real call that pushes a frame; it just returns immediately without recursing. That frame counts, giving five at peak depth.
Heads-up Every call, including the one that hits the base case, pushes a frame. Depth at peak is the number of nested live calls, which is n + 2 here, not the number of recursive steps.
Quiz
Completed
A recursive function is shipped without a reachable base case. What happens at runtime, and why so fast?
Heads-up Without a reachable base case the inputs never satisfy a stopping condition; the descent is unbounded. It does not finish — it crashes when the stack region is exhausted.
Heads-up Most runtimes do not detect infinite recursion or substitute a value. They let the stack grow until the OS signals a stack overflow and terminates the process.
Heads-up Stack frames consume the fixed stack region, not the heap. The failure is a stack overflow against a small reserved range, which is why it happens almost instantly rather than after exhausting all RAM.
Recap
The unit’s through-line is one mechanism seen from several angles: a CALL saves a return address and jumps to a fixed body; each call pushes a LIFO stack frame holding that return address plus the function’s locals; parameters are copied in at CALL and the result handed back at RET; a local’s scope and lifetime are bounded by its frame, so it vanishes on return; and recursion is just the same frame-push repeated, made safe only by a reachable base case — without one, the frames pile up until the stack overflows.