为什么这个功能总是崩溃?

时间:2016-06-11 21:42:59

标签: c recursion

有谁能告诉我为什么fillpool总会崩溃?也许有一个无限的递归,但在哪里?

#include <stdio.h>
#include <stdbool.h>

#define N 5

bool IsNotValidIndex(int Row , int Column)
{
  if((Row >= N || Row < 0) || (Column >= N || Column < 0))
      return true ;
  return false ;
}

void fillpool(int row , int column , int picture[N][N])
{
  if(IsNotValidIndex(row , column))
    return ;
  if(picture[row][column] == 0)
    return ;
  picture[row][column] = 2 ;
  fillpool(row + 1 , column , picture) ;
  fillpool(row - 1 , column , picture) ;
  fillpool(row ,column + 1 , picture) ;
  fillpool(row , column -1 , picture) ;
}

2 个答案:

答案 0 :(得分:1)

您有无限递归,因为您将行/列的值设置为“2”,然后检查它是否为“0”。所以你经常一遍又一遍地将值设置为2。无限递归的发生是因为你正在调用fillpool为“row + 1”,然后它将填充“row-1”,因此你得到无限递归(同样的事情会发生在列+ 1但你永远不会到达那里)

答案 1 :(得分:0)

代码演练可以快速解决问题,通过在调试器中单步执行代码可以轻松完成此过程。

假设您使用row = col = 2来调用它,这会设置2,2

00000
00000
00200
00000
00000

然后你调用fillpool(row + 1 , column , picture),递归直到行无效:

00000
00000
00200
00200
00200

然后调用fillpool(row - 1 , column , picture)来设置1,2

00000
00200
00200
00200
00200

然后调用fillpool(row + 1 , column , picture),但这指的是2,2,因此您正处于开始状态,因此将重复上述序列,直到堆栈耗尽。

4路洪水填充算法因此(来自https://en.wikipedia.org/wiki/Flood_fill):

Flood-fill (node, target-color, replacement-color):
 1. If target-color is equal to replacement-color, return.
 2. If the color of node is not equal to target-color, return.
 3. Set the color of node to replacement-color.
 4. Perform Flood-fill (one step to the south of node, target-color, replacement-color).
    Perform Flood-fill (one step to the north of node, target-color, replacement-color).
    Perform Flood-fill (one step to the west of node, target-color, replacement-color).
    Perform Flood-fill (one step to the east of node, target-color, replacement-color).
 5. Return.

您通过测试0但填充2:

错误地编写了第1步

if(picture [row] [column] == 2)     返回;

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