递归因子

Question

#import <Foundation/Foundation.h>

@interface Factorial : NSObject 
+(int) factorial:(int) n;
@end

@implementation Factorial

+(int) factorial:(int)n
{
    if (n==0) {
        return 1;
    }
    else
    {
        return [self factorial:n]*[self factorial:n-1];
    }
}

@end

int main (int argc, const char * argv[])
{
    int i = [Factorial factorial:5];
    NSLog(@"%d", i);
    return 0;
}

这段代码有什么问题？我是Objective-c的新手（我来自c背景） 或者，我对目标c？

(Compiler generating ..)
GNU gdb 6.3.50-20050815 (Apple version gdb-1708) (Mon Aug  8 20:32:45 UTC 2011)
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sharedlibrary apply-load-rules all
[Switching to process 1413 thread 0x0]
warning: Unable to restore previously selected frame.
(gdb) 

并且在@implementation的行+ +（int）factorial：（int）n中卡住了EXC_BAD_ACESS。

谢谢

Answer 1

您想要计算factorial:n，但在您的函数中使用factorial:n的结果。

变化： return [self factorial:n]*[self factorial:n-1];

于： return n*[self factorial:n-1];

Answer 2

我不熟悉ObjC，但行：

+(int) factorial:(int)n
    //...
    return [self factorial:n]*[self factorial:n-1];
    //                      ^

看起来对我很怀疑。这意味着对于n！= 0，您将获得无限的递归和面堆栈溢出： - ）

Answer 3

您必须更改代码

return n*[self factorial:n-1];

是

return  n* (n != 1 ? [self factorial:n-1] : 1);

第一个将进行无限循环

例如：

int factorial = [self factorial:3];
NSLog(@"factorial 3 >> %i",factorial); the result is "factorial 3 >> 6"