递归函数显示分段问题 C++

Question

此代码用于递归函数练习。当我运行代码时，它停在“POWER” cout 行，然后我的编译器显示分段错误。 POWER 行后面的函数应该递归地将数字“a”提升到数字“b”的幂。我不知道如何解决这个问题，有人可以帮忙吗？

#include <iostream>
#include <string>
#include <vector>
using namespace std;

/**** Recursive backwards print, prints a string starting from last index to first*****/

void printReverse(string s, int i)
{
  if(i < s.size()) 
    {  
      printReverse(s.substr(1), i);
      cout<<s[i];
    }    
    else
    {   
      return;  
    }
  }
 


   /****   Recursive power function, computes a^b, where b can be positive or negative*****/
int recPower(double a, int b)
  {   
    int i = b; //i = b, so int a can be multiplied int b times
     
    if (i == 0) //base 
      return 1;

    else //multiply A by B, B times
     {
       a *= b;
       return recPower(a, b); //recursive
       i--; //decrement i until it equals 0
     }
  }


   
   /****   Recursive string replace, replaces all instances of a character in a string with another character*****/
string recReplace(string s2, int i, char old, char neW)
   { 
       if(s2[i] == old) //search for old char
      {
        i = neW; //replace it
        i++; //iterate i
       }
    recReplace(s2, i, old, neW); //call function
    return s2;
   }



/****   Recursive list find   > Searches if x exists in list, returns true if found, false otherwise*****/
int recListFind(vector<int> v, int i, int x)
  {
    if(v[i] == x)    
      {       
        cout << x << " exists in the vector."<<endl;  
        i++;
        recListFind(v, i, x);
      }    
      return true;
    }



int main()
  {     
    cout << "PRINT REVERSE" << endl;
    cout << "----------" << endl;    
    string s1 = "hello world";    
    cout << "String: " << s1 << endl;    
    cout << "Reversed: ";    
    printReverse(s1, 0);       
    cout << endl;   
     
     
     /* Computes a^b (power function) */    
    cout << "POWER" << endl;    
    cout << "----------" << endl;    
    int a = 2, b = -3;    
    cout << a << "^" << b << " = ";    
    cout << recPower(a, b) << endl;    
    cout << endl;    
   
   
    /* Replaces a character in a string with a new one */    
    cout << "REPLACE" << endl;    
    cout << "----------" << endl;    
    string s2 = "-h-e-l-l-o-";    
    char oldChar = '-';    
    char newChar = ' ';   
    cout << "String: " << s2 << endl;    
    cout << "> Replace '" << oldChar << "' with '" << newChar << endl;    
    recReplace(s2, 0, oldChar, newChar);    
    cout << "String: " <<  s2 << endl;    
    cout << endl;    
   
   
    /* Searches for value in vector */    
    cout << "FIND" << endl;    
    cout << "----------" << endl;    
    int x = 7;    
    cout << "Does " << x << " exist in the vector? ";    vector<int> v = {5, 1, 6, 7, 9};    
    cout << recListFind(v, 0, 7) <<  endl;    
    cout << endl;    
  return 0;
}

Answer 1

问题很简单，您正在使用b 执行recPower 函数。在函数中，如果 b 不为 0，则调用 recPower 时使用未修改的值 b（同时修改 a）。这总是会以无限递归结束，这会溢出你的堆栈。

解决方案可能是：

int recPower(double a, int b, int times) {        
  if (times == 0) 
    return a;

  else
    return b * recPower(a, b, --times); 
}

int recPower(double a, int b) {
    return recPower(a, b, b);
}

即使你解决了这个问题，你还有另一个问题。 b 可以为负数，根据您的逻辑，它将在递减的同时继续递归，直到溢出并返回 0。您将在第一个测试用例中导致这种情况。您应该考虑此函数中允许的类型，考虑将它们设为无符号，或明确处理否定的 b 情况。