通过引用给出的计数器总是等于初始值

Question

我编写了一个程序，通过递归计算给定nuber的阶乘，并另外显示递归调用的数量。

当我将结果存储在分离的变量“result”

时，它按预期工作

long long result = calculateFactorial(factorialNumber, counter);

    cout << "\nfactorial is equal to: " << result
         << "\nRecursion executed: " << counter - 1 << " times.\n";

但是当我想这样做的时候：

    cout << "\nfactorial is equal to: " << calculateFactorial(factorialNumber, counter)
         << "\nRecursion executed: " << counter - 1 << " times.\n";

计数器始终等于初始值-1（-1）。你能解释一下，为什么第二种方法不能正常工作？

代码：

#include <iostream>

using namespace std;

long long calculateFactorial(int factorialNumber, int &counter);

int main() {
    int counter = 0;
    int factorialNumber;

    cout << "Enter factorial number: ";
    cin >> factorialNumber;

    long long result = calculateFactorial(factorialNumber, counter);

    cout << "\nfactorial is equal to: " << result
         << "\nRecursion executed: " << counter - 1 << " times.\n";
    //counter - 1, because first execution is nor recursion.
}

long long calculateFactorial(int factorialNumber, int &counter) {

    if (factorialNumber == 0) {
        return 1;
    }

    if (factorialNumber > 0) {
        counter++;
        return factorialNumber * calculateFactorial(factorialNumber - 1, counter);
    }
    return 2;
    //error code - can be captured
}

谢谢：）