为什么我必须将这些引用参数声明为 const 或按值传递？

Question

#include <vector>
#include <iostream>

using namespace std;

static const int NOT_FOUND = -1;

template <class sequence, class T>
int binarySearch(sequence& seq, int low, int high, T& item)
{
    //..
}

template <class sequence, class T>
int binarySearch(const sequence& seq, const T& item)
{
    if (seq.size() == 0)
        return NOT_FOUND;
    return binarySearch(seq, 0, seq.size() - 1, item);
}

int main()
{
    vector<int> t1 = {0, 3 ,45, 94};
    cout << binarySearch(t1, 0);
    //binarySearch(t1, 0, t1.size() - 1, 45);
    return 0;
}

为什么编译器不接受：

template <class sequence, class T>
int binarySearch(sequence& seq, T& item)

?

此外，为什么程序按规定编译，但调用

binarySearch(t1, 0, t1.size() - 1, 45);

from main 编译不了？

任何情况下的编译器错误都是“没有匹配函数来调用 'binarySearch'。

Answer 1

问题是您不能将临时对象绑定到非常量引用。

binarySearch(t1, 0);
                 ^---- this is a temporary

如果您将 0 存储在一个变量中并使用它，则 non-const version would work.

§ 8.5.4

— Otherwise, if T is a reference type, a prvalue temporary of the type referenced by T is copy-list-initialized or direct-list-initialized, depending on the kind of initialization for the reference, and the reference is bound to that temporary. [ Note: As usual, the binding will fail and the program is ill-formed if the reference type is an lvalue reference to a non-const type. —end note

为什么我必须将这些引用参数声明为 const 或按值传递？

Why do I have to declare these reference paramaters const or pass by value?

c++

templates