函数通过元素指向第一个元素指向的数组

Function to move through an array pointed by the first element by element

本文关键字：元素 数组 第一个 函数      更新时间：2023-10-16

我正在尝试循环遍历数组的元素，对于这些元素，我必须调用一个函数指向的函数，并传递了元素地址，以及在最后一个参数中存储的地址。该功能返回该函数指向返回true的元素数量。这是我一直在尝试遵循的要求，但我无法确切地做我要做的事情。

------要求＆amp;功能------------- 

/* Write an enumeration function named sum() with the following parameters:
    a generic pointer
    an int that holds the number of elements in the array pointed to
    an int that holds the size in bytes of a single element
    a pointer to a function that has two generic pointer parameters and returns a bool
    a generic pointer
Your function moves through the array pointed to by the first parameter element by element.  
For each element, your function calls the function pointed to and passes the element's address    along 
with the address stored in the last parameter.  Your function returns the number of elements for 
which the function pointed to returned true. 
Since your first function parameter is a generic pointer and your function can handle any type, 
you will need to cast the address of the input array to the address of a chars in order to move 
from one element to the next.  */

int sum（void* x，int n，int s，bool（ f）（void ，void*），void* z）{

  char *arr = static_cast<char*>(x);
  int count = 0;
  for (; s < n-2; s++){
      arr += s;
      count += f(arr, z);     
  }
  if (n / 1 == n)
      return count;
  else if (n % 2 == 0)
      return count;
  else
      return 0;

}

我希望你们可以向我展示和解释，或者至少我没有以适当的方式做的事情。.我真的很感激这是唯一的方法，我可以学习.. :)

如果需要更多解释，请告诉我..

** * *** 添加****** *** 

/* Write a callback function named isEven() with the following parameters:
a generic pointer to an input value
a generic pointer to an output value
Your function works with ints and returns true if the input value is even, false otherwise.
Moreover, if the value is even, your function adds the value to that pointed to by the second
parameter. */
bool isEven(void* x, void* z){
  int a = *static_cast<int*>(x);
  int b = *static_cast<int*>(z);
  if (a % 2 == 0){ // finding even numbers
      b += a;
      return true;
  }
  else {
      return false;
  }
}
/* Write another callback function named isPrime() with the following parameters:
a generic pointer to an input value
a generic pointer to an output value
Your function works with ints and returns true if the input value is a prime number, false otherwise.
Moreover, if the value is prime, your function adds the value to that pointed to by the second
parameter. */
bool isPrime(void* x, void* z){
  int a = *static_cast<int*>(x);
  int b = *static_cast<int*>(z);
  if ((a / 1 == a) && (a / a == 1)){ // finding prime numbers
      b += a;
      return true;
  } 
  else {
      return false;
  }
}

** * ** * **** 预期输出* ** * ** * ** ***********

5 evens found in {1,2,3,4,5,6,7,8,9,10,11} sum is 30
5 primes found in {1,2,3,4,5,6,7,8,9,10,11} sum is 28

如果我正确理解了该功能看起来为

int sum( const void* x, int n, int s, bool(*f)( const void*, const void* ), const void* z )
{
   const char *p = reinterpret_cast<const char *>( x );
   int count = 0;
   for ( int i = 0; i < n; i++ )
   {
      count += f( p, z );
      p += s;
   }
   return count;
}

这是使用该函数

的示例 
#include <iostream>
#include <cstdlib>
#include <ctime>
int sum( const void* x, int n, int s, bool(*f)( const void*, const void* ), const void* z )
{
   const char *p = reinterpret_cast<const char *>( x );
   int count = 0;
   for ( int i = 0; i < n; i++ )
   {
      count += f( p, z );
      p += s;
   }
   return count;
}
bool lt( const void *p1, const void *p2 )
{
    return ( *reinterpret_cast<const int *>( p1 ) <
             *reinterpret_cast<const int *>( p2 ) );
}
int main() 
{
    std::srand( ( unsigned int )std::time( 0 ) );
    const int N = 10;
    int a[N];
    for ( int &x : a ) x = std::rand() % N;
    for ( int x : a ) std::cout << x << ' ';
    std::cout << std::endl;
    int x = 5;
    int n = sum( a, N, sizeof( int ), lt, &x );
    std::cout << "There are " << n << " elements less than " << x << std::endl;
    return 0;
}

样本输出

4 7 2 9 2 8 6 1 9 9 
There are 4 elements less than 5

char* arr = (char*) x;
for(; s < (n-1); s++) 
{
    f((void*) &arr[s], z);
    // or
    f((void*) (arr + s * sizeof(char)), z);
}
相关文章：