如何从stdlib为qsort编写比较函数？-Howtowriteacomparefunctionforqsortfromstdlib?

12  

Something like this should work:

像这样的东西应该工作:

int porownaj(const void *p_a, const void *p_b)
{
  /* Need to store arguments in appropriate type before using */
  const pkt *pkt_a = p_a;
  const pkt *pkt_b = p_b;

  /* Return 1 or -1 if alfa members are not equal */
  if (pkt_a->alfa > pkt_b->alfa) return 1;
  if (pkt_a->alfa alfa) return -1;

  /* If alfa members are equal return 1 or -1 if r_kw members not equal */
  if (pkt_a->r_kw > pkt_b->r_kw) return 1;
  if (pkt_a->r_kw r_kw) return -1;

  /* Return 0 if both members are equal in both structures */
  return 0;
}

Stay away from silly tricks like:

远离愚蠢的技巧,如:

return pkt_a->r_kw - pkt_b->r_kw;

which return un-normalized values, are confusing to read, won't work properly for floating point numbers, and sometimes have tricky corner cases that don't work properly even for integer values.

返回非标准化值,令人困惑的读取,对浮点数不能正常工作,有时甚至对于整数值也不能正常工作的棘手角落情况。

1  

There are two parts to the problem - how to write the code, and how to compare the packet types. You must ensure you always return a value. Your code should also always be such that:

该问题分为两部分 - 如何编写代码,以及如何比较数据包类型。您必须确保始终返回值。您的代码也应始终如此:

porownaj(&pkt_a, &pkt_b) == -porownaj(&pkt_b, &pkt_a)

Your outline comparison does not handle cases such as:

您的大纲比较不处理以下情况:

pkt_a->alfa >  pkt_b->alfa && pkt_a->r_kw <= pkt_b->r_kw
pkt_a->alfa alfa && pkt_a->r_kw >= pkt_b->r_kw
pkt_a->alfa == pkt_b->alfa && pkt_a->r_kw != pkt_b->r_kw

There is one more problem - is it appropriate to compare floating point values for exact equality? That will depend on your application.

还有一个问题 - 比较浮点值是否适合精确相等?这取决于您的申请。

Mechanically, you have to convert the const void pointers to const structure pointers. I use the explicit cast - C++ requires it, and I try to make my code acceptable to a C++ compiler even when it is really C code.

在机械上,您必须将const void指针转换为const结构指针。我使用显式强制转换--C ++需要它,我尝试使我的代码可以接受C ++编译器,即使它真的是C代码。

int porownaj(const void *vp1, const void *vp2)
{
     const pkt *pkt_a = (const pkt *)vp1;
     const pkt *pkt_b = (const pkt *)vp2;

     if (pkt_a->alfa >  pkt_b->alfa && pkt_a->r_kw >  pkt_b->r_kw) return 1;
     if (pkt_a->alfa == pkt_b->alfa && pkt_a->r_kw == pkt_b->r_kw) return 0;
     if (pkt_a->alfa alfa && pkt_a->r_kw r_kw) return -1;
     return 0;
 }

This does not deal with the bits that I cannot resolve since I am not party to the necessary information. Note that, in general, multi-dimensional objects (such as complex numbers, or (x,y) or (x,y,z) coordinates) cannot simply be compared for greater than or less than or equal to.

这不涉及我无法解决的比特,因为我不是必要信息的参与者。注意,通常,多维对象(例如复数,或(x,y)或(x,y,z)坐标)不能简单地比较大于或小于或等于。

0  

Yes, I am sorting by alfa and r_kw decides if pkt is first (first value will have the biggest (or smallest) alfa and r_kw I think). That's how I understand the problem, I am not 100% sure.

是的,我正在按照alfa进行排序,而r_kw决定pkt是否是第一个(我认为第一个值将具有最大(或最小)的alfa和r_kw)。这就是我理解这个问题的方法,我不是百分百肯定的。