Why am I getting this error: not all code paths return a value?

可紊 提交于 2019-12-01 14:13:24

The first error, "not all code paths return a value" means there is a path that the code could follow where no value would be returned (ie: calling isHodder(1)). You need to return some value outside of the for loop. Additionally, since you have an if/else block inside the second for loop the line

k=(2^j)-1;

Will never be executed.

    static int isHodder(int n)
    {
        int k = n;
        for (int i = 2; i <= n / 2; i++)
        {
            if ((n % 1) == 0)
            {
                return 0;
            }
            else
            {
                for (int j = 2; j <= k; j++)
                {
                    if (n == (2 ^ j) - 1)
                    {
                        return 1;
                    }
                    else
                    {
                        return 0;
                    }
                    k = (2 ^ j) - 1;
                }
            }
        }
        return 0;
    }
kenny

This line is highly suspect:

if ((n % 1) == 0)

What happens if I call isHodder(1)? Which return statement do I reach?

We may never enter the for loop at all (or if we do enter it, the compiler can't be sure that we reach a return statement before we leave the loop).

@jalf: is correct about the case where 2 <= n/2 (i.e., 4<=n). You will never enter the for loop in that case, so you need the return after the for.

As @Kenny suggested,

if ((n % 1) == 0)

Is suspect. n % 1 always == n, so the condition will only be true when n == 0. However, it also looks like this might be a typo, since the condition does not test anything that varies within the loop. Did you mean

if ((n % i) == 0)

?

For this problem you have to use void if your function didn't return any value otherwise you have to define a return value at the end of your function.

易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!