Question

我知道这里有很多关于递归的问题和答案，但请听我说。在我的程序中，我试图得到一系列只有偶数。用户输入-5000到5000之间的两个数字;如果数字出现故障（即首先输入的数字越大），数字就会被换成顺序;在递归函数中，一个简单的函数返回0或1，使用n%2确定当前值是偶数还是奇数。问题是我无法弄清楚如何实际使用递归。

int recursive_function(int first_number, int second_number)
{

printf("\n    Entering sum function for range %d to %d",
                                                        first_number, second_number);

if(first_number > second_number)
    return 1;

else
{
    if(is_even(first_number) == 0)
    {
        printf("\n        Adding: %d", first_number);
        return (first_number += recursive_function(first_number + 1, second_number));
        printf("\nThis equals added");
    }
    else
    {
        printf("\n        Skipping: %d", first_number);
        return (first_number += recursive_function(first_number + 1, second_number));
        printf("\nThis equals skipped");
    }
}

}

程序工作，就递归终止而言，我只是不明白我应该如何弹出堆栈的其余部分。这是一个应该是输出的表示。（假设值为10和20）

Processing the range 10 to %20:
Entering the sum function for range 10 to 20
    Adding: 10
Entering the sum function for range 11 to 20
    Skipping: 11
Entering the sum function for range 12 to 20
    Adding: 12
Entering the sum function for range 13 to 20
    Skipping: 13
Entering the sum function for range 14 to 20
    Adding: 14
Entering the sum function for range 15 to 20
    Skipping: 15
Entering the sum function for range 16 to 20
    Adding: 16
Entering the sum function for range 17 to 20
    Skipping: 17

...

Entering the sum function for range 20 to 20
    Adding: 20
Entering the sum function for range 21 to 20   (recursion stops here)
(This next part is where I don't know how this happens)
Exiting sum function for range 21 to 20 with result: 0
Exiting sum function for range 20 to 20 with result: 20
Exiting sum function for range 19 to 20 with result: 20
Exiting sum function for range 18 to 20 with result: 38
Exiting sum function for range 17 to 20 with result: 38
Exiting sum function for range 16 to 20 with result: 54

....

Exiting sum function for range 11 to 20 with result: 80
Exiting sum function for range 10 to 20 with result: 90
The sum of all even numbers in the range 10 to 20 is: 90

我知道这与从堆栈中弹出所有这些语句有关。问题是我所做的所有研究都给出了解释，假设一个人真正理解递归，而我是一个业余爱好者。任何人都能用英语帮忙解释一下吗？

Answer 1

要了解发生的事情，请记下一张纸上的一系列电话。使用较小的范围，比如说18到20。

这是一个例子。我省略了大多数函数体留下行与递归调用。

recursive_function(18, 20):                        <<< initial call, depth 0
  printf("\n    Entering
  ...
  printf("\n        Skipping: %d", 18);            <<< Execution goes to this
  return (18 += recursive_function(19, 20)):       <<< recurses, depth 1
    printf("\n    Entering
    ...
    printf("\n        Adding: %d", 19);
    return (19 += recursive_function(20, 20)):     <<< depth 2
        printf("\n    Entering
        ...
        printf("\n        Skipping: %d", 20);
        return (20 += recursive_function(21, 20)): <<< depth 3
            printf("\n    Entering
            ...
            if(21 > 20)
               return 1;                           <<< end of recursion
        printf("\nThis equals skipped");           <<< left over at depth 2
    printf("\nThis equals added");                 <<< left over at depth 1

  printf("\nThis equals skipped");                 <<< left over at depth 0

现在，您可以按顺序收集所有printf语句并获得结果输出。请注意，在您的代码中，您跳过了“退出...”部分，而输出则包含它。

弹出堆栈。弹出堆栈是由编译器为您完成的。它在函数返回时发生。你应该担心在适当的时候返回，而不是如何弹出堆栈。

使用递归c的偶数范围

1 个答案: