Question

我有这个递归函数，它在其素因子中分解一个数字，并显示结果标准输出例如

descompon(2, 10);

输出

2 = 2
3 = 3
4 = 2^2
5 = 5
6 = 2 * 3
7 = 7
8 = 2^3
9 = 3^2
10 = 2 * 5

代码

#include <iostream>
#include <sstream>

int comprobar_primo( int* num, int e )
{
    if (*num%e == 0)
    {
        *num /= e;
        return 1 + comprobar_primo(num, e);
    }
    return 0;
}
std::string factor_primo(int a, int b, std::stringstream& fact)
{
    unsigned exp = comprobar_primo(&a, b);
    if (exp >= 1)
    {
        fact << b;
        if (exp > 1) fact << '^' << exp;
        if (a != 1) fact << " * ";
    }
    if (a > 1) factor_primo(a, b + 1, fact);
    return fact.str();
}
void descompon(int a, int b, int ver)
{
    std::stringstream fact;
    //std::string result = factor_primo(a, 2, fact);
    if(ver)
        std::cout << a << " = " << factor_primo(a, 2, fact) << std::endl;
    if(a < b)
        descompon( a + 1, b, ver);
}
int main(void)
{
    descompon(2, 10000, 1);
    return 0;
}

问题是当达到5922时程序仍然冻结，显示：

Process returned -1073741819 <0xC0000005>

为什么会发生这种情况以及如何避免？

Answer 1

factor_primo和descompon函数都可能导致堆栈溢出。最好将它们转换为迭代版本。修改后的代码如下：

// no need to pass b as argument since we start from b=2 and increment b by 1
std::string factor_primo(int a, std::stringstream& fact)  
{
    for(int b=2; a>1; b++)
    {
        if(a%b==0)
        {
            unsigned exp=comprobar_primo(&a, b);
            if(exp >= 1)
            {
                fact << b;
                if(exp > 1) fact << '^' << exp;
                if(a != 1) fact << " * ";
            }
        }
    }
    return fact.str();
}
void descompon(int a, int b, int ver)
{
    if(ver)
    {
        for(int i=a; i<=b; i++) {
            std::stringstream fact;
            std::cout << i << " = " << factor_primo(i, fact) << std::endl;
        }
    }
}
int main(void)
{
    descompon(2, 10000, 1);
    getchar();
    return 0;
}

分解其主要因素中的数字

1 个答案: