如何将多套件传递到函数

Question

我有一个非常简单的问题。我只是学习地图和多示例，并想知道如何将它们传递到功能中。我的大部分想法都包裹在多图上，但想一个快速的示例，说明如何将它们传递到空白功能中。

int main()
{
multimap<string,int> movies;


movies.insert(pair<string,int>("Happy Feet",6));
movies.insert(pair<string,int>("Happy Feet",4));
movies.insert(pair<string,int>("Pirates of the Caribbean",5));
movies.insert(pair<string,int>("Happy Feet",3));
movies.insert(pair<string,int>("Pirates of the Caribbean",4));
movies.insert(pair<string,int>("Happy Feet",4));
movies.insert(pair<string,int>("Flags of out Fathers",4));
movies.insert(pair<string,int>("Gigli",4));

cout<<"There are "<<movies.count("Happy Feet")<<" instances of "<<"Happy Feet"<<endl;
cout<<"There are "<<movies.count("Pirates of the Caribbean")<<" instances of "<<"Pirates of the Caribbean"<<endl;
cout<<"There are "<<movies.count("Flags of out Fathers")<<" instances of "<<"Flags of out Fathers"<<endl;
cout<<"There are "<<movies.count("Gigli")<<" instances of "<<"Gigli"<<endl;



system("PAUSE");
calculateAverage(movies);  // this is where im getting errors such as no conversions
return 1;
}
void calculateAverage(multimap<string,int> *q)
{
// this function wont calculate the average obviously. I just wanted to test it
int averageH;
int averageP;
int averageF;
int averageG;

averageH = (q->count("Happy Feet"));
averageP = (q->count("Happy Feet"));
averageF = (q->count("Happy Feet"));
averageG = (q->count("Happy Feet"));


};

Answer 1

为什么要通过指针？我认为最好传递参考（如果将地图在函数中修改）或const的参考

void calculateAverage(const multimap<string,int> & q)
{
// this function wont calculate the average obviously. I just wanted to test it
int averageH;
int averageP;
int averageF;
int averageG;

averageH = (q.count("Happy Feet"));
averageP = (q.count("Happy Feet"));
averageF = (q.count("Happy Feet"));
averageG = (q.count("Happy Feet"));
};

Answer 2

通过参考：

void calculateAverage(const multimap<string,int> & q)

但是然后通过指针还不错。只是语法看起来不好。

如果您选择通过指针，则可以在呼叫网站上使用此语法：

calculateAverage(&movies);

Answer 3

在我看来，在我看来，“本着stl的精神”传递给迭代器， movies.begin() 和 movies.end() 到 calculateAverage 功能。例如：

calculateAverage(movies.begin(),movies.end());

并定义了以下定义：

typedef multimap<string,int>::const_iterator MapIt;
void calculateAverage(const MapIt &begin, const MapIt &end)
{
...
}

Answer 4

您正在尝试传递类型的价值 multimap<string,int> 作为指向该类型的指针，即 multimap<string,int>*. 。要么将功能签名更改为 void calculateAverage(const multimap<string,int>& q) 并相应地修改其代码（替换 -> 和 .），或这样称呼： calculateAverage(&movies).