两个程序中的无限递归

Infinite recursion in two programs

本文关键字：无限 递归 程序 两个      更新时间：2023-10-16

我正在写两个程序。一个将一个数提升到另一个数的幂，另一个做最大公约数。它们都使用无限递归崩溃，我不知道为什么。有人能看看这些并给我建议吗？请不要发布完整的解决方案，只发布建议。

#include <iostream>
using namespace std;
int pow( int base, int exp ) {
int somevariable = pow(base,exp-1);
if (base == 0) {
      return 1;
}
else {
      return base * pow (base,exp-1);
      }
}
int main ( ) {
int base;
int power;
cout << "This program calculates exponential values." << endl;
cout << "Enter the base: ";
cin >> base;
cout << "Enter the power: ";
cin >> power;
cout << "" << endl;
cout << base << "^" << power << " =" <<
cout << pow(base, power);
}
#include <iostream>
using namespace std;
int gcd(int number1, int number2) {
int returnj = 0;
if(number1 || number2 >= 0) {
          return gcd(number2, number1 % number2);
           }
           else if(number1 || number2 == 0) {
                return 1;
                }
}
int main ( ) {
int number;
int another;
int gcdd;
cout << "This program calculates the greatest common divisor (GCD) for two integers." << endl;
cout << "Enter a number: ";
cin >> number;
cout << "Enter another: ";
cin >> another;
cout << "" << endl;
cout << "GCD = " << gcd(number, another);

}

在pow函数中，您需要更改

if (base == 0)

收件人：

if (exp == 0)

这样就不会是无休止的递归。

现在可以使用

#include <iostream>
using namespace std;
int pow( int base, int exp ) {
    //There's was a line here creating a loop because it never arrives to a return
    if (exp == 0) { // base 0 makes not sense because the base it's always the same, it's the exp that decreases
          return 1;
    }
    else {
          return base * pow (base,exp-1);
    }
}
int main ( ) {
    int base;
    int power;
    cout << "This program calculates exponential values." << endl;
    cout << "Enter the base: ";
    cin >> base;
    cout << "Enter the power: ";
    cin >> power;
    cout << "" << endl;
    cout << base << "^" << power << " = "; // This line was not closed.
    cout << pow(base, power);
}
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