C++ 递归导致段错误?

C++ recursion causing segfault?

今天我让我的学生写了这段代码,它出现了段错误,我不太明白为什么。问题是为数字的英语发音计算字母。它是最早的欧拉计划问题之一。递归有什么问题吗?

#include<iostream>
using namespace std;

int dlugosc(int x)
{
    if(x == 1) return 3;
    if(x == 2) return 3;
    if(x == 3) return 5;
    if(x == 4) return 4;
    if(x == 5) return 4;
    if(x == 6) return 3;
    if(x == 7) return 5;
    if(x == 8) return 5;
    if(x == 9) return 4;
    if(x == 10) return 3;
    if(x == 11) return 6;
    if(x == 12) return 6;
    if(x == 13) return 7;
    if(x == 14) return 8;
    if(x == 15) return 7;
    if(x == 16) return 7;
    if(x == 17) return 9;
    if(x == 18) return 8;
    if(x == 19) return 8;
    if(x == 20) return 6;
    if(x == 21) return 10;
    if(x == 22) return 10;
    if(x == 30) return 6;
    if(x == 40) return 6;
    if(x == 50) return 5;
    if(x == 60) return 5;
    if(x == 70) return 7;
    if(x == 80) return 6;
    if(x == 90) return 6;
    if(x == 100) return 11;
    if(x == 200) return 11;
    if(x == 300) return 13;
    if(x == 400) return 12;
    if(x == 500) return 12;
    if(x == 600) return 11;
    if(x == 700) return 13;
    if(x == 800) return 13;
    if(x == 900) return 12;
    if(x < 100) return dlugosc(x-x%10) + dlugosc(x%10);
    if(x < 1000) return dlugosc(x-x%100) + dlugosc(x-x%10) + dlugosc(x%10) + 3;
    if(x == 1000) return 11;
}

int main()
{
    int ile = 0;
    for(int i=1; i<=1000; i++)
    {
        ile += dlugosc(i);
    }   
    cout << ile;
    return 0;
}

输入 x=110 不会退出递归。您可以通过 gdb 或手动快速检查:

if(x < 1000) return dlugosc(x-x%100) + dlugosc(x-x%10) + dlugosc(x%10) + 3;

因为 x - x%10x 如果 x 可以被 10 整除,第二项 (dlugosc(x-x%10)) 将无限递归 x = 110.