What is output?

double MyFct(double a, double b) {
   return (a + b) / 2.0;
}

int main() {
   double x = 3.0;
   double y = 5.0;
   double z = 8.0;
   double t;
   t = MyFct(x, y);
   t = MyFct(t, z);
   cout

Question

What is output?

double MyFct(double a, double b) {
   return (a + b) / 2.0;
}

int main() {
   double x = 3.0;
   double y = 5.0;
   double z = 8.0;
   double t;
   t = MyFct(x, y);
   t = MyFct(t, z);
   cout << t << endl;
   return 0;
}

BlackTom AI · Accepted Answer

Function MyFct returns the arithmetic mean of its two double parameters: (a + b) / 2.0.

1) Relevant knowledge points and concepts
- Function call semantics: MyFct(a,b) computes and returns (a+b)/2.0 as a double.
- Order of execution: statements are executed sequentially in main; the result of the first call is stored in t, then t is overwritten by the second call.
- Basic floating-point arithmetic: addition and division by 2.0.

2) Step-by-step reasoning
- Initial values: x = 3.0, y = 5.0, z = 8.0.
- First function call: t = MyFct(x, y).

3) Calculations with intermediate steps
- MyFct(x,y) = (3.0 + 5.0) / 2.0
- = 8.0 / 2.0
- = 4.0
So after the first call, t = 4.0.

- Second function call: t = MyFct(t, z).
- MyFct(4.0, 8.0) = (4.0 + 8.0) / 2.0
- = 12.0 / 2.0
- = 6.0
So after the second call, t = 6.0.

4) Why the correct answer is correct
Because the function computes the average each time, the first average of x and y yields 4.0, and averaging that result with z yields (4.0 + 8.0)/2.0 = 6.0. The final cout prints the double value 6 (which corresponds to 6.0).

5) Why other options are incorrect
- 4.0: This is the intermediate value after the first call, but the program makes a second call that changes t to 6.0.
- 16.0: No operation in the code produces 16.0 (this would be 8.0+8.0 or a product, neither performed).
- 8.0: This is the value of z, but t is overwritten by the average of t and z; the final result is not simply z.

6) Conclusion
The final printed value is 6.0, matching the provided correct answer.

What is output? double MyFct(double a, double b) { return (a + b) / 2.0; } int main() { double x = 3.0; double y = 5.0; double z = 8.0; double t; t = MyFct(x, y); t = MyFct(t, z); cout << t << endl; return 0; }单项选择题

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