C++:不能将泛型函数作为参数传递给另一个函数

Question

我想将多个比较函数传递给选择排序函数,如下所示,但我得到了以下错误:

Error   1   error C2664: 'sort' : cannot convert parameter 3 from 'bool (__cdecl *)(int,int)' to 'bool *(__cdecl *)(T,T)'   c:\users\milad\documents\visual studio 2008\projects\functionpass\functionpass\passcompare.cpp  49  FunctionPass

代码:

bool integerCompare (int a , int b)
{
    return(a<b);
}
bool charCompare (char a , char b)
{
    return(a<b);
}
bool stringCompare (string a , string b)
{
    if(a.compare(b)<0) return true;
    else return false;
}
template <class T>
void sort(T x[], int n , bool(*whichCompare(T,T))) // n=size of the array
{
    for (int pass=0; pass<n-1; pass++) {
        int potentialSmallest = pass;  
        for (int i=pass+1; i<n; i++) {
            if ((*whichCompare)(x[i],x[potentialSmallest])) {
                potentialSmallest = i;
            }
        }

        int temp = x[pass];
        x[pass] = x[potentialSmallest];
        x[potentialSmallest] = temp;
    }
}
template <typename T>
void printArray(T a[], int size)
{
    for(int i=0;i<size;i++)
        cout<<" "<<a[i];
}
int main()
{
    int intArray[] = {1,7,-8,-14,46,33,4};
    sort <int>(intArray , 7 , integerCompare);
    printArray<int>(intArray,7);
}

Answer 1

你有这个:

template <class T> void sort(T x[], int n , bool(*whichCompare(T,T)))
{ /*...*/ }

返回a bool和两个类型参数的函数指针的参数声明T是错误的.你可能真的想要:

template <class T> void sort(T x[], int n , bool (*whichCompare)(T,T))
{ /*...*/ }

虽然通常这样的函数是这样编写的:

template <class T, typename Functor>
void sort(T x[], int n , Functor whichCompare)
{
    // ...
    bool result = whichCompare(x[i], x[potentialSmallest]);
    // ...
}

这样,用户不仅可以传递函数指针,还可以传递提供以下内容的函数对象operator()():

struct MyIntegerCompareFunctor
{
    bool operator()(int a, int b) { return a < b; }
};

sort(intArray, 7, MyIntegerCompareFunctor());
sort(intArray, 7, &integerCompare); // Works too

C++标准库提供的一些算法就像这样编写.