C动态内存分配和sizeof() [英] C dynamic memory allocation and sizeof()

问题描述

我在理解这两个代码段之间的区别时遇到了一些麻烦: 我使用以下语句在代码内为整数数组动态分配空间

I'm having some trouble understanding the difference between these two code segments: I allocate space for an array of integers dynamically within my code with the following statement

int *arr = calloc(cnt, sizeof(int));

在另一个函数中，我要传入arr，我想确定arr的大小(元素数). 当我打电话

In another function, where I pass in arr, I would like to determine the size (number of elements) in arr. When I call

int arr_sz = sizeof(arr)/sizeof(int);

它仅返回1，这只是我假设的两个参数(4/4)= 1的int中的字节数. 我只是假设这与使用数组相同

it only returns 1, which is just the number of bytes in an int for both arguments I am assuming (4/4)=1. I just assumed it would be the same as using an array

  int arr[8];
  int arr_sz = sizeof(arr)/sizeof(int);

返回数组中元素的实际数量.

which returns the actual number of elements in the array.

如果有人可以解决这个问题，那就太好了.谢谢！

If anyone could clear this up, that would be great. Thanks!

推荐答案

int *arr;   ----> Pointer
int arr[8]; ----> Array

首先要了解的内容-int *arr是一个指针，指向某些字节的内存位置，不是一个数组.

First up what you got there - int *arr is a pointer, pointing to some bytes of memory location, not an array.

数组和指针的类型不相同.

The type of an Array and a Pointer is not the same.

在传递arr的另一个函数中，我想确定arr的大小(元素个数).当我打电话

int arr_sz = sizeof(arr)/sizeof(int);

它仅返回1，这只是我假设的两个参数(4/4)= 1的int中的字节数.我只是认为这与使用数组相同

即使假定是成为一个数组，这也是因为数组在传递给函数时会衰减为指针.您需要在函数中显式传递数组大小作为单独的参数.

Even if it is assumed to be an Array -- that's because Arrays get decayed into pointers when passed into functions. You need to explicitly pass the array size in functions as a separate argument.

经历一下:

使用C编程语言编写的数组的大小?

static数组与动态内存分配之间存在差异.

There is a difference between a static array and dynamic memory allocation.

sizeof运算符不适用于动态分配. AFAIK最适合与基于堆栈的类型和预定义类型一起使用.

The sizeof operator will not work on dynamic allocations. AFAIK it works best with stack-based and predefined types.

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