Discussion
Consider the following program:
int main()
{
f1();
f2(2);
f3();
return(0);
}
int f1()
{
return(1);
}
int f2(int X)
{
f3();
if (X==1)
return f1();
else
return (X*f2(X-1));
}
int f3()
{
return(5);
}
Which one of the following options represents the activation tree corresponding to the main function?
- A.
- B.
- C.
- D.
Answer: A
The activation tree is (A).
An activation tree has one node per call, with each call's children in the order it makes them. Trace the program:
- main calls f1(), then f2(2), then f3(): three children, in that order.
- f2(2) first calls f3(); since X = 2 it then calls f2(1).
- f2(1) first calls f3(); since X = 1 it then calls f1() and returns.
So main has children f1, f2, f3; that f2 has children f3 and f2; and the inner f2 has children f3 and f1. This is tree (A).
(B) stops after one level of f2 and misses the recursive call f2(1). (C) makes f2 a child of f1, but f1 calls nothing. (D) puts f3, f2 and f1 all under the first f2, when f1 is called by the inner f2(1).
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