小编ceo*_*eod的帖子

为什么以下代码返回分段错误？当我注释掉第7行时,seg错误消失了.

int main(void){
      char *s;
      int ln;
      puts("Enter String");
      // scanf("%s", s);
      gets(s);
      ln = strlen(s); // remove this line to end seg fault
      char *dyn_s = (char*) malloc (strlen(s)+1); //strlen(s) is used here as well but doesn't change outcome
      dyn_s = s;
      dyn_s[strlen(s)] = '\0';
      puts(dyn_s);
      return 0;
    }

干杯!

下面是两个定义的函数,用于查找数字列表的最大值.

mx :: (Ord a) => [a] -> a
mx [] = error "Empty list"
mx [x] = x
mx (x:xs) 
    | x > (mx xs) = x
    | otherwise = (mx xs)

mx' (x:xs) = findMax x xs
  where
    findMax cmx [] = cmx
    findMax cmx (x:xs) | x > cmx = findMax x xs
               | otherwise = findMax cmx xs

main = do 
    print $ mx [1..30]

定时上面的代码,首先是mx'(尾递归),然后是mx(非尾递归),我们有以下时序.

Lenovo-IdeaPad-Y510P:/tmp$ time ./t 
30

real    0m0.002s
user    0m0.000s
sys 0m0.001s …