为什么std :: sort崩溃如果比较函数不是作为运算符<? [英] Why will std::sort crash if the comparison function is not as operator <?
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问题描述
使用VC ++ 2012编译以下程序。
The following program is compiled with VC++ 2012.
#include <algorithm>
struct A
{
A()
: a()
{}
bool operator <(const A& other) const
{
return a <= other.a;
}
int a;
};
int main()
{
A coll[8];
std::sort(&coll[0], &coll[8]); // Crash!!!
}
如果我更改返回一个< = other。 a;
to return a<
If I change return a <= other.a;
to return a < other.a;
then the program runs as expected with no exception.
为什么?
推荐答案
std :: sort
需要一个满足 strict weak ordering 规则的分拣机,解释
此处
std::sort
requires a sorter which satisfies the strict weak ordering rule, which is explained
here
所以,你的对手说那个 a< b
当 a == b
不遵循严格弱排序规则时,算法有可能将会崩溃,因为它将进入无限循环。
So, your comparer says that a < b
when a == b
which doesn't follow the strict weak ordering rule, it is possible that the algorithm will crash because it'll enter in an infinite loop.
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