这个简单的c ++程序可以永远运行，我不知道为什么

Question

It compiles fine, prints the first "start" but it stops right there. Any help is greatly appreciated. I spent several hours trying to figure out whats wrong and tried running it within several different IDEs. I think it fails at the while-loop.

#ifndef TERNARY_SEARCH_H
#define TERNARY_SEARCH_H

#include <cstdlib>
#include <iostream>


template <typename ArrayLike, typename T>
int ternary_search(const ArrayLike& array, const T& value, int low, int high) 
{
/* 
 * low is the lowest possible index, high is the highest possible index
 * value is the target value we are searrching for
 * array is the ascending order array we are searching 
 */


bool found = false; 

while(!found)
{
    int lowerThirdIndex =   (((high - low)/(3)) + low);
    int upperThirdIndex = (2*((high - low)/(3)) + low);

// search lower third
    if (array[lowerThirdIndex] == value) 
    {
      return lowerThirdIndex;
      found = true;
    }
    else if (array[lowerThirdIndex] > value)
    {
        high = lowerThirdIndex;
    }
    else // array[lowerThirdIndex] < value
    {
        low = lowerThirdIndex;
    }

    //search upper third
    if (array[upperThirdIndex] == value) 
    {
      return upperThirdIndex;
      found = true;
    }
    else if (array[upperThirdIndex] > value)
    {
        high = upperThirdIndex;
    }
    else // array[upperThirdIndex] < value
    {
        low = upperThirdIndex;
    }

}
  return -1;
}


#endif /* TERNARY_SEARCH_H */


//main.cpp
#include "ternary_search.h"
using namespace std;

int main() {
  cout << "start";
  int nums[] = {0, 10, 20, 30, 40, 50, 60, 70, 80, 90};
  for (int i = 0; i <= 90; i += 10) {
    if (ternary_search(nums, i, 0, 10) != i / 10) {
      std::cout
        << "Searching for " << i << " returned index "
        << ternary_search(nums, i, 0, 10) << " instead of "
        << i / 10 << "." << std::endl;
      return 1;
    }

    // search for something that doesn't exist.
    if (ternary_search(nums, i + 1, 0, 10) != -1) {
      std::cout
        << "Searching for " << i + 1 << " returned index "
        << ternary_search(nums, i + 1, 0, 10) << " instead of -1."
        << std::endl;
      return 1;
    }
  }
  std::cout << "On this small example, your search algorithm seems correct.\n";
  return 0;
}

Answer 1

Your ternary_search function doesn't have a means to return when it fails to find the value in the search table. It returns only when it finds an element in the table that exactly matches the value you pass in.

Since the second invocation of the function is called with i+1 -- which is 1 -- which is not a member of your table, your ternary search function never exits.