c++ - Armstrong 计算函数崩溃

Question

我正在用 C++ 编写一个函数，计算所有阿姆斯特朗数字，直到作为参数发送给该函数的上限。出于某种原因，我的函数没有将 153 识别为 armstrong 号码，并且在识别 9474 后崩溃。这是我目前所拥有的:

void isArmstrong(int curr_value, int arm) {

    //tmp value to hold original upper limit
    int curr_value1 = curr_value;

    //accumulator to keep track of how many armstrong numbers have been found
    int armstrong = 0;

    //determining how many digits long the current tested number is for armstrong testing
    string currStr = to_string(curr_value);

    //the power the current tested number will be raised to
    int nth = currStr.length();

    //cout <<"num " << armStr<<endl<<"nth "<<nth << endl;

    //variable to break out of loop
    int accum = 0;

    //sum variable to check if the number is armstrong after computational testing
    int sum = 0;

    //temp number to add to the variable 'sum'
    int tmp = 0;

    //cout <<"-----"<< curr_value << endl;

    while (accum <= nth) {
        //calculating a numbers sum by raising each digit to the nth power and adding that to sum
        int digit = curr_value1 % 10;
        curr_value1 = (curr_value1 / 10);

        tmp = pow(digit, nth);

        sum += tmp;

        accum += 1;

    }

    //if sum and the current value are the same, the number is an armstrong number
    if (sum == curr_value) {
        armstrong += 1;
        cout << sum << endl;
    }

    //making sure the current value is less than the upper limit and calling the function again
    if (curr_value < arm) {
        isArmstrong(curr_value += 1, arm);
    }

}

下面是我如何从 main 调用函数:

isArmstrong(1, 54748);

这是控制台输出

1
2
3
4
5
6
7
8
9
370
371
407
1634
8208
9474

Answer 1

与其进行太多有可能使您的堆栈变满的递归调用，不如这样做。

for(int i=1;i<=54748;i++) if(check_armstrong(i)){//..do your work };

简单地说，我的建议是进行迭代调用，而不是进行太多的递归调用。

过多的递归调用会使您的程序停止，因为堆栈已被调用函数的帧填满。这就是为什么您不应该依赖于深度递归，而应该使用独立于堆栈内存的迭代版本。

每次调用函数时，您都需要在内存中存储有关该函数的一些数据。因此只有一定数量的调用可以放入内存中。在这里，当您调用函数 9474 次时，内存已满。这就是它停止的原因。

加分项:

1. 需要检查条件。 while(accum<nth) .

2. 您将 armstrong 数字的数量存储在局部变量中。如果你想使用它，你可以使用全局变量(不推荐)或者你可以简单地将它放在函数之外并与函数本身分开使用。

函数应该是模块化的，只执行一个 Action 。在我建议的解决方案中，您可以看到，您可以重复使用 check-armstrong函数多次，这是使用函数的理想本质。

提示:

bool check_armstrong(int curr_value) {

    //tmp value to hold original upper limit
    int curr_value1 = curr_value;


    //determining how many digits long the current tested number is for armstrong testing
    string currStr = to_string(curr_value);

    //the power the current tested number will be raised to
    int nth = currStr.length();

    //variable to break out of loop
    int accum = 0;

    //sum variable to check if the number is armstrong after computational testing
    int sum = 0;

    //temp number to add to the variable 'sum'
    int tmp = 0;


    while (accum < nth) {
        //calculating a numbers sum by raising each digit to the nth power and adding that to sum
        int digit = curr_value1 % 10;
        curr_value1 = (curr_value1 / 10);

        tmp = pow(digit, nth);

        sum += tmp;

        accum += 1;

    }

    //if sum and the current value are the same, the number is an armstrong number
    if (sum == curr_value) {
        return true
    }
    return false;

}

在循环中你可以这样做

for( int i =1;i<=... )
{
   if( check_armstrong(i))
   {
      cout<<i<<endl;
      armstrong_count++;
   }
}