AP CSP Day 31: Procedure Abstraction
Share
Practice Question
What is displayed after the following code runs?
p ← 2
q ← 5
p ← p + q
q ← p - q
p ← p - q
DISPLAY(p)
DISPLAY(q)This is the classic swap-without-a-temporary-variable algorithm. Trace: p=2, q=5. After p ← p+q: p=7. After q ← p-q: q=7-5=2. After p ← p-q: p=7-2=5. Final values: p=5, q=2. The values are swapped.
A) The values are exchanged, not preserved. C) p is modified again in the third step — students who stop tracing after line 4 get stuck on p=7. D) q is updated in the second step to 2, not kept at 5.
Students stop tracing too early, reporting p=7 after the third line and forgetting that p is reassigned again. Every assignment must be traced to completion.
When a variable is reassigned multiple times in sequence, always use the most recently updated value. Never skip a step or use a stale value.
Keep Practicing!
Consistent daily practice is the key to AP CSP success.
AP CSP Resources Get 1-on-1 Help