双指针作为函数的数组 [英] Double pointer as an array to a function

问题描述

这是我程序的示例代码.

This is a sample code of my program.

我有一个名为 function 的函数，该函数用于返回整数值和整数数组.我必须将它们作为指针传递(不能更改 function 的签名).所以我写了下面的代码.在这里，我必须将值分配给传递给该函数的双指针.

I have a function named function which is used to return an integer value and an integer array. I have to pass them as pointers (The signature of function cannot be changed). So I wrote the below code. Here I have to assign the values to the double pointer passed to the function.

void function(unsigned int* count, unsigned int ** array)
{
  for(unsigned int i = 0; i<4;i++)
  {
    //array[i] = (unsigned int*)i*10;
  }
  *count = 4;
}

int main()
{

  unsigned int count;
  unsigned int array2[4];
  function(&count, (unsigned int**)&array2);

  return 0;
}

但是上面的代码不起作用.

But the above code is not working.

推荐答案

通过重用概念，我想您已经知道，可以使用指向数组第一个元素的指针，然后使用将指针的地址传递给函数.

By reusing concepts I suppose you already know, you might use a pointer to the first element of the array and then pass the pointer's address to the function.

这样，您将有一个双指针指向一个无符号整数，该整数是数组的第一个元素:

In that way you would have a double pointer to an unsigned int which is the first element of the array:

void function(unsigned int* count, unsigned int ** array)
{
  for(unsigned int i = 0; i<4;i++)
  {
    (*array)[i] = i*10; // WATCH OUT for the precedence! [] has higher precedence!
  }
  *count = 4;
}

int main(int argc, char* argv[])
{

  unsigned int count;
  unsigned int array2[4];
  unsigned int *pointer_to_array = &array2[0];
  function(&count, (unsigned int**)&pointer_to_array);

  return 0;
 }

---

作为旁注:

---

As a sidenote:

如果您可以更改签名，这将更有意义:

If you could change the signature this would have made more sense:

void function(unsigned int* count, unsigned int * array)
{
  for(unsigned int i = 0; i<4;i++)
  {
    array[i] = i*10;
  }
  *count = 4;
}

int main(int argc, char* argv[])
{

  unsigned int count;
  unsigned int array2[4];
  function(&count, array2);

  return 0;
 }

以上方法起作用的原因是，当您将数组传递给函数时(直接或使用指向该数组的显式指针)，它实际上成为了一个指针.这称为阵列衰减.

The reason the above works is because when you pass an array into a function (directly or with an explicit pointer to that array) it essentially becomes a pointer. This is called array decaying.

这篇关于双指针作为函数的数组的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！