Base CS from zero
Variables and state: multiple-choice review
Crux Multiple-choice synthesis across the variables unit — binding vs value, assignment as a store, mutation and state, copy vs reference semantics, and aliasing traps.
Your altitude — climbing toward senior
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You are at middle altitude — in the skySix questions that cut across the whole unit. Each one is a decision about what is actually stored in a cell — a value or a reference — and what changes when you assign or mutate. Get the memory model right and aliasing bugs stop being surprises.
Goal
Confirm you can connect the four ideas the unit built up: a variable is a name bound to a cell, assignment overwrites that cell, mutation changes state in place, and primitives copy while objects share by reference.
Quiz
Completed
After 'let score = 0', the program runs 'score = 42'. What changed in memory?
Heads-up A name is a source-code label; it never holds a value and never 'becomes' a number. The name 'score' stays 'score' — the cell it is bound to now contains 42.
Heads-up Re-assignment writes in place into the existing cell. It does not allocate a new cell or change the binding — only the cell's contents change.
Heads-up Assignment is a destructive overwrite. The previous value 0 is gone; the cell holds only 42. Memory keeps no history of past values.
Quiz
Completed
Why is reading '=' as 'equals' a trap, and what should you read it as instead?
Heads-up As an equation 'x = x + 1' is impossible. It is not an equation at all; it is a two-step command — compute x+1, then store it back — so it is read 'x gets x+1'.
Heads-up Comparison is a different operator (== or ===). A single '=' performs a write, not a test. It always stores the right-hand value into the left-hand cell.
Heads-up TypeScript evaluates eagerly. The right-hand side is computed to a concrete value immediately, and that value is written into the cell now — there is no deferred evaluation.
Quiz
Completed
A bug report says 'I changed obj2 and obj1 mutated on its own.' The code was 'let obj2 = obj1'. What actually happened?
Heads-up There is no ongoing sync. Assignment copied the address once; afterwards both variables independently hold the same address, which points to a single object. There were never two objects.
Heads-up Objects are not deep-copied on assignment — only the reference is copied. The single underlying object is shared, which is exactly why the mutation appeared on both names.
Heads-up It is the defining behaviour of aliasing. Both names hold the same reference to one object, so 'obj2.field = x' is the same write that 'obj1.field' reads back.
Quiz
Completed
'const user = { name: "A" }; user.name = "B";' runs without error. Yet 'const x = 1; x = 2;' throws. Why?
Heads-up const is enforced for objects too — rebinding 'user = somethingElse' still throws. What it does not do is prevent mutation of the object's properties; it only seals which reference the cell holds.
Heads-up Objects are values; the variable's cell holds a reference value. const seals that reference against replacement, but the heap object behind it is not frozen, so property mutation is fine.
Heads-up Whether the property exists is irrelevant. const guards the binding, not the contents; adding a new property (user.age = 1) would also be allowed.
Quiz
Completed
'let a = [1,2]; let b = a; b.push(3);' — then code elsewhere reads a.length expecting 2. Why does it see 3?
Heads-up push mutates the existing array in place; it does not create a new array or rebind anything. Because a and b alias that array, the new length is visible through a.
Heads-up let vs const is irrelevant to sharing. The arrays are shared because the reference was copied on 'let b = a'; the declaration keyword does not change that.
Heads-up Arrays are objects in JavaScript, not primitives. Assigning one copies its reference, so both names point at a single array — only primitives (number, string, boolean) copy by value.
Quiz
Completed
What is the difference between a program's 'state' and the 'mutations' that produced it?
Heads-up State is the current snapshot, not the history. Reaching score=10 by one big increment or ten small ones gives different histories but identical state — only the present values matter.
Heads-up State covers every live variable, including ones untouched for a long time. Their current values are still part of the program's condition right now.
Heads-up They are distinct: state is the values held now; mutation is the act of changing one of those values in place. You can have unchanging state (only const) with no mutation at all.
Recap
The through-line: a variable is a name bound to a cell, and the name and the value are different things. Assignment is a destructive store — ‘gets’, not ‘equals’. Mutation changes a cell in place, and state is the snapshot of all current cell contents. The decisive split is what sits in the cell: a primitive value (copied on assignment, so changes do not propagate) or a reference to a heap object (copied as an address, creating aliases where one mutation is visible through every name). const seals which reference a cell holds, never the contents behind it.