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将指向数组的指针传递给函数(C ++)

[英]Passing pointer to an array into a function (C++)

I'm trying to pass an array into my function calls for build_max_heap and max_heapify so I can modify the array after each call, but I receive an error saying "candidate function not viable: no known conversion from 'int [9]' to 'int *&'for 1st argument." 我正在尝试将数组传递给我的函数调用build_max_heap和max_heapify,所以我可以在每次调用后修改数组,但是我收到一条错误,说“候选函数不可行:没有已知的从'int [9]'转换为' int *&'为第一个参数。“

#include <iostream>
#include <string>
using namespace std;

void build_max_heap(int*& array, int size);
void max_heapify(int*& array, int size, int index);


void build_max_heap(int*& array, int size)
  {
      for(int i = size/2; i>=0; i--)
      {
          max_heapify(array, i);
      }
  }


void max_heapify(int*& array, int size, int index)
  {
      int leftChild = 2*index+1;
      int rightChild = 2*index+2;
      int largest;
      int heap_size = size;

      if( leftChild <= heap_size && array[leftChild] > array[index])
          largest = leftChild;
      else
          largest = index;

      if(rightChild <= heap_size && array[rightChild] > array[largest])
          largest = rightChild;

      if(largest != index) {
          int tempArray = array[index];
          array[index] = array[largest];
          array[largest] = tempArray;
          max_heapify(array, heap_size, largest);
      }

  }

int main()
{
      int array[]={5,3,17,10,84,19,6,22,9};
      int size = sizeof(array)/sizeof(array[0]);

      build_max_heap(array, size);

      return 0;
}

int array[]={5,3,17,10,84,19,6,22,9};

While array can be decayed to a pointer int* to be passed as a function argument, it the pointer cannot be passed as a "non-const reference" int*& , because it is immutable (it is a constant address). 虽然array可以衰减为指针int*以作为函数参数传递,但指针不能作为“非const引用” int*&传递,因为它是不可变的(它是一个常量地址)。 You could have passed it as a const reference like this: 您可以将它作为const引用传递,如下所示:

void max_heapify(int* const& array, int size, int index)
//                    ^^^^^^

However, this doesn't make much sense, you can simply pass the pointer by value (a copy of the address of the array), which results in the same: the variable at the caller wont be changed. 但是,这没有多大意义,您可以简单地按值传递指针(数组地址的副本),这会导致相同的结果:调用者的变量不会被更改。 The usual use case of const& parameters is to pass objects that are expensive to copy, such as std::string . const&参数的通常用例是传递复制成本高的对象,例如std::string This does not apply to pointers; 这不适用于指针; making a copy of a pointer is as cheap as copying any basic variable. 制作指针的副本与复制任何基本变量一样便宜。

You should change your functions to take the pointer by value: 您应该更改函数以按值获取指针:

void build_max_heap(int* array, int size)
void max_heapify(int* array, int size, int index)

also, correct the call to max_heapify inside build_max_heap , give it the correct number of arguments: 同时,纠正调用max_heapifybuild_max_heap ,给它正确的参数个数:

void build_max_heap(int* array, int size)
{
   for(int i = size/2; i>=0; i--)
   {
       max_heapify(array, size, i);  // <-- 3 arguments
   }
}

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