How many function calls exist in the following code?

int Calc1 (int a, int b) {
   return a + b / 2;
}
int Calc2 (int a, int b) {
   return a * b / 100;
}
int main () {
   int x;
   int y;
   x = Calc1 (5,3);
   cout

Question

How many function calls exist in the following code?

int Calc1 (int a, int b) {
   return a + b / 2;
}
int Calc2 (int a, int b) {
   return a * b / 100;
}
int main () {
   int x;
   int y;
   x = Calc1 (5,3);
   cout << x;
   y = Calc2 (5,3);
   cout << y;
   cout <<
   Calc2 (5,3);
}

BlackTom AI · Accepted Answer

Calc1(5,3) is invoked once and Calc2(5,3) is invoked twice, so there are 3 calls to the user-defined functions in main.

Relevant concepts:
- A function call is an execution occurrence of a function (e.g., Calc1(5,3) causes Calc1 to run).
- A function definition (the code that defines Calc1 or Calc2) is not a call.
- Built-in or overloaded operators such as cout << are implemented as functions in C++, but in typical counting problems like this we count explicit calls to the user-defined functions named in the code (Calc1 and Calc2).

Step-by-step reasoning with each line that matters:
1. x = Calc1(5,3); — This executes Calc1 once. Count = 1.
2. cout << x; — This prints x. It does not call Calc1 or Calc2. Count remains 1.
3. y = Calc2(5,3); — This executes Calc2 once. Count = 2.
4. cout << y; — Prints y. No Calc1/Calc2 call. Count remains 2.
5. cout << Calc2(5,3); — This evaluates Calc2(5,3) to obtain a value to send to cout, so Calc2 is called again. Count = 3.

Intermediate details:
- Definitions:
  int Calc1(int a, int b) { return a + b / 2; } — defines the function; not a call.
  int Calc2(int a, int b) { return a * b / 100; } — defines the function; not a call.
- There are no nested function calls inside the return expressions; a + b/2 and a*b/100 use arithmetic only.

Why the correct answer is correct:
- Only three explicit executions of the user-defined functions occur in main: one Calc1 and two Calc2 calls, totaling 3.

Why other options are incorrect:
- 5: would overcount by treating unrelated operations (or counting both operator<< overloads) as additional user-defined function calls.
- 2: would count only unique function names (Calc1 and Calc2) rather than actual invocations.
- 4: is one too many; there are only two Calc2/one Calc1 calls, not four.

Final confirmation: The total number of function calls is 3, matching the provided answer "3".

How many function calls exist in the following code? int Calc1 (int a, int b) { return a + b / 2; } int Calc2 (int a, int b) { return a * b / 100; } int main () { int x; int y; x = Calc1 (5,3); cout << x; y = Calc2 (5,3); cout << y; cout << Calc2 (5,3); }单项选择题

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