无法读取此递归函数 [英] Can't read this recursive function

问题描述

大家好

我正在使用这段代码。

我不知道怎么读它，尤其是'|| 在函数的两个召唤之间。

谢谢。

Hi guys
i'm having truoble with this code.
i dont know how to read it, especially the ' || ' between the two summonus of the function.
thanks.

#include <stdio.h>

int f3(int *a, int n, int m);
int f4(int *a, int n);
void f5(int *a, int n);

int main()
{
	int arr[6] = {3, 12, 6, 11, 2, 10};
	char *txt[] = {"NO","YES","MAYBE","PROBABLY"};

	printf("%s\n",txt[f3(arr,6,5)]);
	return 0;
}
int f3(int *a, int n, int m)
{
	if(m==0) return 1;
	if(n<2) return 0;
	return f3(a+1,n-1,m-a[0]) || f3(a+1,n-1,m);
}

推荐答案

除了上述内容之外，以下修改可能有助于更好地理解该问题。我相信'enum'在C和C ++中的作用相同。

In addition to the above, perhaps the following reworking helps better understand the issue. I believe 'enum' works the same in C as C++.

#include <stdio.h>

enum Branch { InitialEntry, LeftBranch, RightBranch };
int f3(int *a, int n, int m, Branch b);
 
int main()
{
	int arr[6] = {3, 12, 6, 11, 2, 10};
	char *txt[] = {"NO","YES","MAYBE","PROBABLY"};
 
	printf("%s\n",txt[f3(arr,6,5, InitialEntry)]);
	return 0;
}
int f3(int *a, int n, int m, Branch b)
{
   if(b==InitialEntry) printf("%s %d %d\n", "Entry ",n, m);
   else if(b==LeftBranch) printf("%s %d %d\n", "Left ",n, m);
   else printf("%s %d %d\n", "Right ",n, m);
	if(m==0) return 1;
	if(n<2) return 0;
	return f3(a+1,n-1,m-a[0], LeftBranch) || f3(a+1,n-1,m, RightBranch);
}

|| 运算符是逻辑 OR 运算符，如 http：// msdn中所述。 microsoft.com/en-us/library/x04xhy0h(v=VS.80).aspx [ ^ ]。在上面的代码中， return 语句将返回两次调用 f3 函数的逻辑和。至于这个功能实际上做了什么，我没有分析它，所以不能告诉你。

The || operator is the logical OR operator as described in http://msdn.microsoft.com/en-us/library/x04xhy0h(v=VS.80).aspx[^]. In the above code the return statement will return the logical sum of the two calls to the f3 function. As to what this function actually does, I have not analysed it so cannot tell you.

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