Question

我尝试使用C ++中的STL创建类型为Job的最小优先级队列。我有以下代码：

class Job
{
public:
    int IOtime;
};

struct grThan
{
    bool operator()(const Job& l, const Job& r) const
    {  
        return l.IOtime > r.IOtime;  
    }  
};

我正在声明队列std::priority_queue<Job, vector<Job>, grThan> IO;

我收到以下错误：

bool std::operator <(const std::basic_string<_Elem,_Traits,_Alloc> &,const _Elem *)' : could not deduce template argument for 'const std::basic_string<_Elem,_Traits,_Alloc> &' from 'const Job'

Answer 1

// stackoverflow_26691997.cpp (cX) 2015 adolfo.dimare@gmail.com
// http://stackoverflow.com/questions/26691997/

#include <queue>
#include <vector>

class Job {
public:
    int IOtime;
};

struct grThan {
    bool operator()(const Job& l, const Job& r) const {
        return l.IOtime > r.IOtime;
    }
};

int main() {
    std::priority_queue<Job, std::vector<Job>, grThan> IO;
}

带对象的最小优先级队列

1 个答案: