以下程序使用VC++ 2012编译.
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!!!
}
如果我将 return a <= other.a;
更改为 return a <other.a;
然后程序按预期运行,无一例外.
If I change return a <= other.a;
to return a < other.a;
then the program runs as expected with no exception.
为什么?
std::sort
需要一个满足严格弱排序规则的排序器,具体说明这里
std::sort
requires a sorter which satisfies the strict weak ordering rule, which is explained
here
所以,你的比较器说 a <b
当a == b
不遵循严格弱排序规则时,算法可能会崩溃,因为它会进入无限循环.
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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