-5 
#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objective c的概念有误？？

(Compiler generating ..)
GNU gdb 6.3.50-20050815 (Apple version gdb-1708) (Mon Aug  8 20:32:45 UTC 2011)
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This GDB was configured as "x86_64-apple-darwin".tty /dev/ttys002
sharedlibrary apply-load-rules all
[Switching to process 1413 thread 0x0]
warning: Unable to restore previously selected frame.
(gdb)

并在@implementation导致EXC_BAD_ACESS处卡在+（int）阶乘：（int）n行。

谢谢

4

3 回答 3

2

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

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

到： return n*[self factorial:n-1];

于 2012-12-10T07:44:42.220 回答
1

我不熟悉 ObjC，但行：

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

看起来对我很可疑。这意味着对于 n != 0 您将获得无限递归和面对堆栈溢出 :-)

于 2012-04-29T07:15:03.103 回答
1

你必须改变你的代码

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

成为

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"
于 2016-07-23T12:58:39.657 回答