std :: sort与自定义比较器 [英] std::sort with custom comparator
问题描述
在下面的代码中,为什么所有三个 IntComparator()
, IntComparator2
和 IntComparator3
都充当的第三个参数sort()
函数?它们不会具有不同的l值函数类型吗?基于 https://en.cppreference.com/w/cpp/algorithm/sort它说
In the following code, why do all three IntComparator()
, IntComparator2
and IntComparator3
work as the 3rd parameter of the sort()
function? Wouldn't they have different l-value function types? Based on https://en.cppreference.com/w/cpp/algorithm/sort it says
比较功能的签名应等效于以下:
The signature of the comparison function should be equivalent to the following:
bool cmp(const Type1& a,const Type2& b);
bool cmp(const Type1 &a, const Type2 &b);
哪个似乎更匹配 IntComparator2
?
还有哪一个更可取?第三种选择似乎更简单,更直观.
Also which one would be preferable? Third option seems much simpler and more intuitive.
#include <iostream>
#include <vector>
#include <algorithm>
#include <functional>
struct IntComparator
{
bool operator()(const int &a, const int &b) const
{
return a < b;
}
};
bool IntComparator2 (const int &a, const int &b)
{
return a < b;
}
bool IntComparator3 (int a, int b)
{
return a < b;
}
int main()
{
int items[] = { 4, 3, 1, 2 };
std::sort(items, items+4, IntComparator());
for (int n=0; n<4; n++) {
std::cout << items[n] << ", ";
}
std::cout << "\n";
int items2[] = { 4, 3, 1, 2 };
std::sort(items2, items2+4, IntComparator2);
for (int n=0; n<4; n++) {
std::cout << items2[n] << ", ";
}
std::cout << "\n";
int items3[] = { 4, 3, 1, 2 };
std::sort(items3, items3+4, IntComparator3);
for (int n=0; n<4; n++) {
std::cout << items3[n] << ", ";
}
std::cout << "\n";
return 0;
}
推荐答案
std :: sort
接受 functor
.这是可以调用的任何对象(带有正确的参数).该功能通过使用以下模板来实现此目的
std::sort
accepts a functor
. This is any object that can be called (with the correct parameters). The function achieves this by using templates, like the following
template<typename Iter, typename Comp>
void sort(Iter begin, Iter end, Comp compare) { ... }
IntComparator1
,2和3都是此比较器的有效函子,因为它们都可以使用带有2个整数的operator()进行调用.
IntComparator1
, 2, and 3 are all valid functors for this comparator, since they can all be called using operator() with 2 integers.
也像您说的那样,第三个选项通常确实更直观.
Also like you said, the third option is indeed usually more intuitive.
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