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C puzzles详解【34-37题】

2024-07-24 01:15:45   216人阅读

第三十四题

The following is a piece of C code, whose intention was to print a minus sign 20 times. But you can notice that, it doesn‘t work.   #include <stdio.h>  int main()  {      int i;      int n = 20;      for( i = 0; i < n; i-- )          printf("-");      return 0;  }Well fixing the above code is straight-forward. To make the problem interesting, you have to fix the above code, by changing exactly one character. There are three known solutions. See if you can get all those three.

题目讲解：

将

for( i = 0; i < n; i-- )

改成

for( i = 0; i < n; n-- )

 

第三十五题

What‘s the mistake in the following code?   #include <stdio.h>  int main()  {      int* ptr1,ptr2;      ptr1 = malloc(sizeof(int));      ptr2 = ptr1;      *ptr2 = 10;      return 0;  }

题目讲解：

int* ptr1,ptr2;

ptr1是指针，ptr2不是指针

改成

int *ptr1,*ptr2;

 

第三十六题

What is the output of the following program?   #include <stdio.h>  int main()  {      int cnt = 5, a;      do {          a /= cnt;      } while (cnt --);      printf ("%d\n", a);      return 0;  }

题目讲解：

cnt减到最后为0，运行后有“trap divide error”“floating point exception”的错误。

 

第三十七题

What is the output of the following program?   #include <stdio.h>  int main()  {      int i = 6;      if( ((++i < 7) && ( i++/6)) || (++i <= 9))          ;      printf("%d\n",i);      return 0;  }

题目解答：

i的值为8。先执行(++i < 7)，此表达式的值为0，i=7，由于逻辑运算符的短路处理，(i++/6)跳过执行，
((++i < 7) && ( i++/6))值为0，接着执行(++i <= 9)，i的值最终为8。

C puzzles详解【34-37题】


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